Answer: 0.373

P(Chester comes from Dream island) = # of penguins in dream island / # of all penguins in the data = \frac{55+68}{330} \approx 0.373

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: 0.447

P(Chester is an Adelie penguin given that Chester comes from Dream island) = # of Adelie penguins from Dream island / # of penguins from Dream island = \frac{55}{55+68} \approx 0.447

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Difficulty: ⭐️

The average score on this problem was 91%.

Answer: 0.575

Method 1

P(Chester is not an Adelie penguin given that Chester is not from Dream island) = # of penguins that are not Adelie penguins from islands other than Dream island / # of penguins in island other than Dream island = \frac{119\ \text{(eliminate all penguins that are Adelie or from Dream island, only Gentoo penguins from Biscoe are left)}}{44+44+119} \approx 0.575

Method 2

P(Chester is not an Adelie penguin given that Chester is not from Dream island) = 1- (# of penguins that are Adelie penguins from islands other than Dream island / # of penguins in island other than Dream island) = 1-\frac{44+44}{44+44+119} \approx 0.575

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Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: 0.027

For all three flights to be on United, we need the first flight to be on United, and the second, and the third. Since these are independent events that do not impact one another, and we need all three flights to separately be on United, we need to multiply these probabilities, giving an answer of 0.3*0.3*0.3 = 0.027.

Note that on an exam without calculator access, you could leave your answer as (0.3)^3.

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Difficulty: ⭐️

The average score on this problem was 93%.

Answer: 0.099

We already calculated the probability of all three flights being on United as (0.3)^3 = 0.027. Similarly, the probability of all three flights being on Delta is (0.4)^3 = 0.064, and the probability of all three flights being on American is (0.2)^3 = 0.008. Since we cannot satisfy more than one of these conditions at the same time, we can separately add their probabilities to find a total probability of 0.027 + 0.064 + 0.008 = 0.099.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: False

It’s not quite the same because the previous subpart doesn’t include the probability that all three flights are on the same airline which is not one of Delta, United, or American. For example, there is a small probability that all three flights are on Allegiant or all three flights are on Southwest.

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Difficulty: ⭐️

The average score on this problem was 90%.

Answer: 0.027

For all three flights to be on United, we need the first flight to be on United, and the second, and the third. Since these are independent events that do not impact one another, and we need all three flights to separately be on United, we need to multiply these probabilities, giving an answer of 0.3*0.3*0.3 = 0.027.

Note that on an exam without calculator access, you could leave your answer as (0.3)^3.

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Difficulty: ⭐️

The average score on this problem was 93%.

Answer: 0.099

We already calculated the probability of all three flights being on United as (0.3)^3 = 0.027. Similarly, the probability of all three flights being on Delta is (0.4)^3 = 0.064, and the probability of all three flights being on American is (0.2)^3 = 0.008. Since we cannot satisfy more than one of these conditions at the same time, we can separately add their probabilities to find a total probability of 0.027 + 0.064 + 0.008 = 0.099.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: False

It’s not quite the same because the previous subpart doesn’t include the probability that all three flights are on the same airline which is not one of Delta, United, or American. For example, there is a small probability that all three flights are on Allegiant or all three flights are on Southwest.

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Difficulty: ⭐️

The average score on this problem was 90%.

Answer: \left( 1 - \frac{1}{7}\right)^7 = \left( \frac{6}{7}\right)^7

Of the 7 rows in the seven_apps DataFrame, there are 6 rows that have a value less than 3 in the dependents column. This means that if we were to sample one row from seven_apps, there would be a \frac{6}{7} chance of selecting one of the rows that has less than 3 dependents. The question is asking what the probability that the maximum number of dependents in the resample is less than 3. One resample of the DataFrame is equivalent to sampling one row from seven_apps 7 different times, without replacement. So the probability of getting a row with less than 3 dependents, 7 times consecutively, is \left( \frac{6}{7}\right)^7.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 47%.

Answer: \left[ \left( 1 - \frac{1}{7}\right)^7 \right]^{50} = \left( \frac{6}{7}\right)^{350}

We know from the previous part of this question that the probability of one resample of seven_apps having a maximum number of dependents less than 3 is \left( \frac{6}{7}\right)^7. Now we must repeat this process 50 times independently, and so the probability that all 50 resamples have a maximum number of dependents less than 3 is \left(\left( \frac{6}{7}\right)^{7}\right)^{50} = \left( \frac{6}{7}\right)^{350}. Another way to interpret this is that we must select 350 rows, one a time, such that none of them are the one row containing 3 dependents.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 41%.

Answer: \left[1 - \left( 1 - \frac{1}{7}\right)^7 \right]^{50} = \left[1 - \left( \frac{6}{7}\right)^7 \right]^{50}

We’ll first take a look at the probability of one resample of seven_apps having the maximum number of dependents be 3. In order for this to happen, at least one row of the 7 selected for the resample must be the row containing 3 dependents. The probability of getting this row at least once is equal to the complement of the probability of never getting this row, which we calculated in part (a) to be \left( \frac{6}{7}\right)^7. Therefore, the probability that at least one row in the resample has 3 dependents, is 1 -\left( \frac{6}{7}\right)^7.

Now that we know the probability of getting one resample where the maximum number of dependents is 3, we can calculate the probability that the same thing happens in 50 independent resamples by multiplying this probability by itself 50 times. Therefore, the probability that the maximum number of dependents is 3 in each of 50 resamples is \left[1 - \left( \frac{6}{7}\right)^7 \right]^{50}.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 27%.

Answer: a = 11, b = 12

The format of the answer suggests we should use the complement rule. The opposite of at least one application being approved is that no applications are approved, or equivalently, all applications are denied.

Consider one application. Its probability of being approved is \frac{3}{6}*\frac{1}{6} = \frac{3}{36} = \frac{1}{12} because we need to get any one of the three even numbers on the first roll, then the second roll must match the first. So one application has a probability of being denied equal to \frac{11}{12}.

Therefore, the probability that all k applications are denied is \left(\frac{11}{12}\right)^k. The probability that this does not happen, or at least one is approved, is given by 1-\left(\frac{11}{12}\right)^k.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 41%.

Answer: \frac{1}{36}

Imagine generating a security code with three unique digits by selecting one digit at a time. In other words, we would need to select three values without replacement from the set of digits 0, 1, 2, \dots, 9. The probability that the first digit is odd is \frac{5}{10}. Then, assuming the first digit is odd, the probability of the middle digit being 0 is \frac{1}{9} since only nine digits are remaining, and one of them must be 0. Then, assuming we have chosen an odd number for the first digit and 0 or the middle digit, there are 8 remaining digits we could select, and only 4 of them are even, so the probability of the third digit being even is \frac{4}{8}. Multiplying these all together gives the probability that all three criteria are satisfied: \frac{5}{10} \cdot \frac{1}{9} \cdot \frac{4}{8} = \frac{1}{36}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 38%.

Answer: \frac{1}{40}

We’ll use a similar strategy as in the previous part. This time, however, we need to select three values with replacement from the set of digits 0, 1, 2, \dots, 9. The probability that the first digit is odd is \frac{5}{10}. Then, assuming the first digit is odd, the probability of the middle digit being 0 is \frac{1}{10} since any of the ten digits can be chosen, and one of them is 0. Then, assuming we have chosen an odd number for the first digit and 0 or the middle digit, the probability of getting an even number for the third digit is \frac{5}{10}, which actually does not depend at all on what we selected for the other digits. In fact, when we sample with replacement, the probabilities of each digit satisfying the given criteria don’t depend on whether the other digits satisfied the given criteria (in other words, they are independent). This is different from the previous part, where knowledge of previous digits satisfying the criteria informed the chances of the next digit satisfying the criteria. So for this problem, we can really just think of each of the three digits separately and multiply their probabilties of meeting the desired criteria: \frac{5}{10} \cdot \frac{1}{10} \cdot \frac{5}{10} = \frac{1}{40}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

Answer: Audi

Let’s compute the number of EVs that are either SUVs or sedans for each non-Nissan Brand. (To do this, we’ll look at the right-most two columns in the DataFrame provided.)

• Number of Teslas that are SUVs or sedans: 4 + 3 = 7
• Number of BMWs that are SUVs or sedans: 1 + 1 = 2
• Number of Audis that are SUVs or sedans: 8 + 1 = 9

Since Audi is the "Brand" with the most total SUVs and sedans, it is the most likely "Brand" to be selected.

Note: You could compute conditional probabilities for each brand, if you’d like, by dividing the counts above by 18 (the total number of SUVs and sedans). For instance, P(\text{EV is a BMW given that EV is an SUV or sedan}) = \frac{2}{18}. The "Brand" with the highest count (Audi, with 9 SUVs or sedans) is also the "Brand" with the highest conditional probability of being selected given that the selected car is an SUV or sedan (Audi, with \frac{9}{18}).

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Difficulty: ⭐️⭐️

The average score on this problem was 88%.

Answer: 92

In the first selection, the probability of selecting a BMW is \frac{1+1+1}{24} = \frac{3}{24} (3 is the total number of EVs that are BMW, and 24 is the total number of non-Nissan EVs as given by the question).

In the second selection, since we select without replacement, there are only 23 EVs we can select from. Given that in the first selection we already selected 1 BMW, there are only 2 BMWs left among the 23 EVs left. Thus, the probability of getting a BMW in the second selection is \frac{2}{23}.

Putting this all together, the probability that both selections are BMWs is

\frac{3}{24}\cdot\frac{2}{23} = \frac{6}{24} \cdot \frac{1}{23} =\frac{1}{4} \cdot \frac{1}{23} = \frac{1}{92}

So, k = 92.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: \frac{4}{13}

The question is asking for the proportion of SUVs that are made by Tesla.

We first need to find the number of SUVs in the DataFrame provided, which is 4 + 1 + 8 = 13. Of those 13 SUVs, 4 are made by Tesla. Thus, the proportion of SUVs made by Tesla is \frac{4}{13}, so the probability that a randomly selected SUV is made by Tesla is \frac{4}{13}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 66%.

Answer: 0.7

We can simply count the number of transactions meeting at least one of the two criteria. More easily, there are only 3 rows that do not meet either of the criteria (the rows that are "visa" and "credit" transactions). Therefore, the probability is 7 out of 10, or 0.7. Note that we cannot simply add up the probability of "mastercard" (0.3) and the probability of "debit" (0.6) since there is overlap between these. That is, some transactions are both "mastercard" and "debit".

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: \frac{2}{15}

We know that the sample space here doesn’t have any of the $100 transactions, so we can ignore the first 4 rows when calculating the probability. In the remaining 6 rows, there are exactly 2 debit transactions with "visa" cards, so the probability of our first transaction being of the specified type is \frac{2}{6}. There are also two credit transactions with "visa" cards, but the denominator of the probability of the second transaction is 5 (not 6), since the sample space was reduced by one after the first transaction. We’re choosing without replacement, so you can’t have the same transaction in your sample twice. Thus, the probability of the second transaction being a visa credit card is \frac{2}{5}. Now, we can apply the multiplication rule, and we have that the probability of both transactions being as described is \frac{2}{6} \cdot \frac{2}{5} = \frac{4}{30} = \frac{2}{15}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: \left(\frac{3}{10}\right)^{15}

There are only 3 rows in the sample with a transaction amount under $25, so the chance of choosing one transaction with such a low value is \frac{3}{10}. For the maximum transaction amount to be less than 25 dollars, this means all transaction amounts in our sample have to be less than 25 dollars. To find the chance that all transactions are for less than $25, we can apply the multiplication rule and multiply the probability of each of the 15 transactions being less than $25. Since we’re choosing 15 times with replacement, the events are independent (choosing a certain transaction on the first try won’t affect the probability of choosing it again later), so all the terms in our product are \frac{3}{10}. Thus, the probability is \frac{3}{10} * \frac{3}{10} * \ldots * \frac{3}{10} = \left(\frac{3}{10}\right)^{15}.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: \frac{3}{5}

Given that the chosen individual is from a family with one child, we know that they must be from either Family X or Family Z. There are three individuals in Family X, and there are a total of five individuals from these two families. Thus, the probability of choosing any one of the three individuals from Family X out of the five individuals from both families is \frac{3}{5}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: False

If two events are independent, knowledge of one event happening does not change the probability of the other event happening. In this case, events A and B are not independent because knowledge of one event gives complete knowledge of the other.

To see this, note that the probability of choosing a child randomly out of the fifteen individuals is \frac{9}{15}. That is, P(B) = \frac{9}{15}.

Suppose now that we know that the chosen individual is an adult. In this case, the probability that the chosen individual is a child is 0, because nobody is both a child and an adult. That is, P(B \text{ given } A) = 0, which is not the same as P(B) = \frac{9}{15}.

This problem illustrates the difference between mutually exclusive events and independent events. In this case A and B are mutually exclusive, because they cannot both happen. But that forces them to be dependent events, because knowing that someone is an adult completely determines the probability that they are a child (it’s zero!)

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

Answer: True

If two events are independent, the probability of one event happening does not change when we know that the other event happens. In this case, events C and D are indeed independent.

If we know that the chosen individual is a child, the probability that they come from Family Y is \frac{3}{9}, which simplifies to \frac{1}{3}. That is P(D \text{ given } C) = \frac{1}{3}.

On the other hand, without any prior knowledge, when we select someone randomly from all fifteen individuals, the probability they come from Family Y is \frac{5}{15}, which also simplifies to \frac{1}{3}. This says P(D) = \frac{1}{3}.

In other words, knowledge of C is irrelevant to the probability of D occurring, which means C and D are independent.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: p^5

The probability of selecting Family W in the first round is p, which is the same for the second round, the third round, and so on. Each of the chosen letters is drawn independently from the others because the result of one draw does not affect the result of the next. We can apply the multiplication rule here and multiply the probabilities of choosing Family W in each round. This comes out to be p\cdot p\cdot p\cdot p\cdot p, which is p^5.

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Difficulty: ⭐️

The average score on this problem was 91%.

Answer: 1 - (1 - p)^5

Since there are too many ways that Family W can be selected to meet the condition that it is selected at least once, it is easier if we calculate the probability that Family W is never selected and subtract that from 1. The probability that Family W is not selected in the first round is 1-p, which is the same for the second round, the third round, and so on. We want this to happen for all five rounds, and since the events are independent, we can multiply their probabilities all together. This comes out to be (1-p)^5, which represents the probability that Family W is never selected. Finally, we subtract (1-p)^5 from 1 to find the probability that Family W is selected at least once, giving the answer 1 - (1-p)^5.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: p \cdot (1 - p)^4

We want to find the probability of Family W being selected only as the last draw, and not in the first four draws. The probability that Family W is not selected in the first draw is (1-p), which is the same for the second, third, and fourth draws. For the fifth draw, the probability of choosing Family W is p. Since the draws are independent, we can multiply these probabilities together, which comes out to be (1-p)^4 \cdot p = p\cdot (1-p)^4.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: (1 - p)^{10} * p^2

The probability of February and December being romance is p for each of those two months, and the probability of any one of the other months being non-romance is 1-p. We multiply all twelve of these probabilities since the months are independent, and the result is p^2(1−p)^{10}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: 1 - (1 - p)^4

To solve this, we calculate the probability that none of the last four months contains romance, which is (1−p)^4. Taking the complement gives us the probability that at least one of the last four months is romance: 1−(1−p)^4. The two possible situations are: no romance and at least one romance, so together they must add up to one, which is why we can use the complement.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.

Answer: p^{12} + (1 - p)^{12}

In this case, there are only two scenarios to consider: either all 12 books are non-romance, which gives (1−p)^{12}, or all 12 books are romance, which gives p^{12}. Adding these together gives p^{12}+(1−p)^{12}.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 29%.

Answer: \frac{(n - 1)}{n} * \frac{(n - 2)}{n} * \frac{(n - 3)}{n}

For there to be no duplicate books in the first four months, each subsequent book must be different from the previous ones. After selecting the first book (which will always be undiplicated), the second book has \frac{(n - 1)}{n} probability of being unique, the third \frac{(n - 2)}{n}, and the fourth \frac{(n - 3)}{n}. If you like, you can think of the probability of the first book being unique as \frac{n}{n} and include it in your answer.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 31%.

Answer: j/i

Looking at the code, we can see that i represents the number of books with a rating greater than 3, and j represents the number of books with a rating greater than 4. This is a conditional probability question: find the probability that Michelle’s book has a rating greater than 4 given that it has a rating greater than 3. Since we already know Michelle’s book has a rating higher than 3, the denominator, representing the possible books she could get, should be i. The numerator represents the books that meet our desired condition of having a rating greater than 4, which is j. Therefore, the result is j/i.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

Although i represents the total number of books with a rating greater than 3 and k represents the number of romance books, there is no information about how many romance books have a rating greater than 3. Thus we cannot find the answer without additional information.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

Answer: \dfrac{40}{100} \cdot \dfrac{39}{99} \cdot \dfrac{38}{98}

We need to find the probability that we get all Twix among the three candies selected from the bowl. Since we are selecting three times from the same bowl, we know that we are selecting without replacement.

1. First Selection:
   • There are 40 Twix and 40 + 50 + 10 = 100 candies in the bowl, meaning the probability of selecting a Twix is \frac{40}{100}.
2. Second Selection:
   • Now that we have chosen one Twix there are 39 Twix and 99 candies left, meaning that the probability of selecting a Twix now is \frac{39}{99}.
3. Third Selection:
   • After selecting two Twix there are 38 Twix and 98 candies left, meaning the probability of selecting a Twix is \frac{38}{98}.

The total probability that we grab all Twix from the bowl is the product of these probabilities: \frac{40}{100} \cdot \frac{39}{99} \cdot \frac{38}{98}

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: \dfrac{60}{100} \cdot \dfrac{59}{99} \cdot \dfrac{58}{98}

We need to find the probability that we get no Twix among the three candies selected from the bowl. We know that two candies are not Twix in our bowl (M&Ms and Kitkats). Since we are selecting three times from the same bowl, we know that we are selecting without replacement.

1. First Selection:
   • There are 60 non-Twix candies in the bowl (50 M&Ms and 10 Kitkats) and 100 total candies. This means the probability of selecting a non-Twix is \frac{60}{100}.
2. Second Selection:
   • Regardless of which non-Twix candy was chosen, there are now 59 non-Twix candies in the bowl (49 M&Ms and 10 Kitkats OR 50 M&Ms and 9 Kitkats). Since there are 99 total candies in the bowl, the probability of selecting a non-Twix is \frac{59}{99}.
3. Third Selection:
   • After selecting two non-Twix there are 58 non-Twix and 98 total candies left meaning the probability of selecting a non-Twix is \frac{58}{98}.

The total probability that we grab no Twix from the bowl is the product of these probabilities: \frac{60}{100} \cdot \frac{59}{99} \cdot \frac{58}{98}

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: 1 - a - b or 1 - (a + b)

The case where we get some Twix and some non-Twix occurs can also be thought of as the case when we DO NOT get either all Twix OR all non-Twix. In 6.1 we calculated the probability of getting all Twix as a and in 6.2 we calculated the probability of getting all non-Twix as b. Therefore the probability of getting either all Twix OR all non-Twix is equal to a + b. However, we are looking for the probability that this does not happen, meaning our answer is 1 - (a + b).

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 30%.

Answer: 0.2

We want to find the probability that Stella assembles the bed frame in 10 minutes, given that she assembles it in less than 30 minutes. The multiplication rule can be rearranged to find the conditional probability of one event given another.

\begin{aligned} P(A \text{ and } B) &= P(A \text{ given } B)*P(B)\\ P(A \text{ given } B) &= \frac{P(A \text{ and } B)}{P(B)} \end{aligned}

Let’s, therefore, define events A and B as follows:

• A is the event that Stella assembles the bed frame in 10 minutes.
• B is the event that Stella assembles the bed frame in less than 30 minutes.

Since 10 minutes is less than 30 minutes, A \text{ and } B is the same as A in this case. Therefore, P(A \text{ and } B) = P(A) = 0.1.

Since there are only two ways to complete the bed frame in less than 30 minutes (10 minutes or 20 minutes), it is straightforward to find P(B) using the addition rule P(B) = 0.1 + 0.4. The addition rule can be used here because assembling the bed frame in 10 minutes and assembling the bed frame in 20 minutes are mutually exclusive. We could alternatively find P(B) using the complement rule, since the only way not to complete the bed frame in less than 30 minutes is to complete it in exactly 30 minutes, which happens with a probability of 0.5. We’d get the same answer, P(B) = 1 - 0.5 = 0.5.

Plugging these numbers in gives our answer.

\begin{aligned} P(A \text{ given } B) &= \frac{P(A \text{ and } B)}{P(B)}\\ &= \frac{0.1}{0.5}\\ &= 0.2 \end{aligned}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: 0.4

We are told that the time it takes someone to assemble the bedside table is a random quantity, independent of the time it takes them to assemble the bed frame. Therefore we can disregard the information about the time it took him to assemble the bed frame and read directly from the probability distribution that his probability of assembling the bedside table in 40 minutes is 0.4.

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Difficulty: ⭐️⭐️

The average score on this problem was 82%.

Answer: 0.53

There are several different ways for the total assembly time to take at most 60 minutes:

1. The bed frame takes 10 minutes and the bedside table takes any amount of time.
2. The bed frame takes 20 minutes and the bedside table takes 30 or 40 minutes.
3. The bed frame takes 30 minutes and the bedside table takes 30 minutes.

Using the multiplication rule, these probabilities are:

1. 0.1*1 = 0.1
2. 0.4*0.7 = 0.28
3. 0.5*0.3 = 0.15

Finally, adding them up because they represent mutually exclusive cases, we get 0.1+0.28+0.15 = 0.53.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 58%.

Answer: (1-p)^n

If you go alone, it means all of your friends failed to come. We can think of this as an and condition in order to use multiplication. The condition is: your first friend doesn’t come and your second friend doesn’t come, and so on. The probability of any individual friend not coming is 1-p, so the probability of all your friends not coming is (1-p)^n.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: 1-p^n

It’s actually easier to figure out the opposite event. The opposite of somebody no-showing is everybody shows up. This is easier to calculate because we can think of it as an and condition: the first artist shows up and the second artist shows up, and so on. That means we just multiply probabilities. Therefore, the probability of all artists showing up is p^n and the probability of some artist not showing up is 1-p^n.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: [-4, -3. -2, -1]

Let’s break down the code line by line:

• .groupby("Major") will group the survey DataFrame by the "Major" column.
• .count() will count the number of students in each major.
• .sort_values("College") sort these counts in ascending order using the "College" column (which now contains counts).
• .take([-4, -3. -2, -1]) will take the last 4 rows of the DataFrame, which will correspond to the 4 most common majors.
• .get("Section") get the "Section" column, which also contains counts.
• .plot(kind="barh") will plot these counts in a horizontal bar chart.

The argument we give to take is [-4, -3, -2, -1], because that will give us back a DataFrame corresponding to the counts of the 4 most common majors, with the most common major ("MA30") at the bottom. We want the most common major at the bottom of the Series that .plot(kind="barh") is called on because the bottom row will correspond to the top bar – remember, kind="barh" plots one bar per row, but in reverse order.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 4%.

Answer: \frac{1}{4}

• Total number of students in popular majors: 30 + 25 + 15 + 10 = 80
• The number of students with "EN" in their major code is the sum of "EN30" and "EN25" students: 25 + 15 = 40

The probability that a single student chosen at random from the popular majors has "EN" in their major code is number of "EN" majors divided by total number of popular majors.

P(\text{one has ``EN"}) = \frac{40}{80} = \frac{1}{2}

Since the students are chosen with replacement, the probabilities remain consistent for each draw. So the probability that both randomly chosen students (with replacement) have "EN" in their major code is

P(\text{both have ``EN"}) = P(\text{one has ``EN"}) \cdot P(\text{one has ``EN"}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 61%.

Answer: \frac{3}{32}

• The probability of selecting one "CG35" major first and then one "MA30" major is: \frac{10}{80} \cdot \frac{30}{80} = \frac{1}{8} \cdot \frac{3}{8} = \frac{3}{64}
• The probability of selecting one "MA30" major first and then one "CG35" major is: \frac{30}{80} \cdot \frac{10}{80} = \frac{3}{8} \cdot \frac{1}{8} = \frac{3}{64}

Since the two events are mutually exclusive (they cannot both happen), we sum their probabilities to get the combined probability of the event occurring in any order:
P(\text{one ``CG35" and one ``MA30"}) = P(\text{``CG35" then ``MA30" OR ``MA30" then ``CG35"}) = P(\text{``CG35 then ``MA30"}) + P(\text{``MA30" then ``CG35"}) = \frac{3}{64} + \frac{3}{64} = \frac{6}{64} = \frac{3}{32}

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 18%.

Answer: \frac{8^k - 7^k}{8^k}

Of the 80 students in popular majors, 10 are in "CG35". To determine the likelihood of selecting at least one "CG35" major out of k random draws with replacement, we first find the probability of not drawing a "CG35" major in all k attempts by following the steps below:

• The probability of not selecting a "CG35" major in one draw is 1 - \frac{10}{80} = 1 - \frac{1}{8} = \frac{7}{8}

• The probability of not selecting a "CG35" major in k draws is, then, \frac{7^k}{8^k}, since each draw is independent.

Now, by subtracting this probability from 1, we get the probability of drawing at least one "CG35" major in those k attempts. So, the probability of selecting at least one "CG35" major in k attempts is 1 - \frac{7^k}{8^k} = \frac{8^k}{8^k} - \frac{7^k}{8^k} = \frac{8^k - 7^k}{8^k}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 58%.

Answer: \displaystyle\frac{1}{5}

From Kathleen’s perspective, there are 5 tutors that are equally likely to become her roommate. So, the probability that Sophia ends up being Kathleen’s roommate is \displaystyle\frac{1}{5}.

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Difficulty: ⭐️

The average score on this problem was 90%.

Answer: \frac{1}{15}

In order to get this combination of roommates, we can use similar logic as before. From Kathleen’s perspective, there are 5 tutors that are equally likely to become her roommate. So, the probability that Sophia ends up being Kathleen’s roommate is \displaystyle\frac{1}{5}. From there, we can view the situation from Kate’s perspective; Kate sees that there are 3 potential roommates left (Ashley, Claire, Vivian). So, the probability that Sophia ends up being Kathleen’s roommate (given Kathleen and Sophia are together) is \displaystyle\frac{1}{3}. After Kate chooses her roommate, Claire and Vivian end up together by process of elimination. We can multiply these two probabilities to recieve: \displaystyle\frac{1}{5} \cdot \displaystyle\frac{1}{5} = \displaystyle\frac{1}{15}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 47%.

Answer: (contacts.shape[0])/(10**10)

• contacts.shape[0]: retrieves the number of rows in the contacts DataFrame, representing the total number of phone numbers you have stored in your contacts.
• 10**10: Since the phone number consists of ten digits, and each digit can be any number from 0 to 9, there are a total of 10**10 possible ten-digit numbers.

Given that each digit pressed is equally likely and independent of others, the likelihood of hitting a specific number from your contacts by random chance is simply the count of your contacts divided by the total number of possible combinations (which is 10**10).

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: \dfrac{1}{4 \cdot 3 \cdot 2 \cdot 1}

The cat has already dialed “987-654”. Since the first six digits are fixed and chosen without replacement, the only remaining digits to be dialed are 3, 2, 1, and 0. The sequence “3210” must be dialed in that exact order from the remaining digits.

• The probability of dialing ‘3’ first is \dfrac{1}{4} (4 digits are remaining).
• The probability of then dialing ‘2’ next is \dfrac{1}{3} (3 digits are remaining).
• The probability of then dialing ‘1’ next is \dfrac{1}{2} (2 digits are remaining).
• The probability of lastly dialing ‘0’ is \dfrac{1}{1} (as it’s the only digit left).

The product of these probabilities gives \dfrac{1}{4 \cdot 3 \cdot 2 \cdot 1}, representing the likelihood of this specific sequence occurring.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 55%.

Answer: \frac{0.14}{0.36}=\frac{7}{18}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 53%.

Answer: False

If two events A and B are independent then P(A and B) = P(A)P(B). In this question we can define A as “selecting a shipping container from Vietnam” and B as “selecting a shipping container from Ireland”. Therefore (A and B) would be “selecting a (single) shipping container from both Vietnam and Ireland”. Since that is impossible, P(A and B) = 0 while P(A)P(B) = 0.012. Therefore these two events are not independent.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 32%.

Answer: 0.03

The probability of selecting Germany as the first shipping container is 0.05. Given that Germany is selected first, the probability that the second container is from the United States is 0.3. However, since the order of the containers can be swapped, you need to multiply by 2 to account for both possible arrangements.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

Answer: 0.59^{3}

There are three important pieces of information to take note:

• Three shipping containers
• Replacement
• None of them come from Asia

This means that we have three trials with replacement, all looking for the same outcome. Given that the probability of a shipping container is .41, we know that the complement is:

\text{Not from Asia} = 1 - .41 \text{Not from Asia} = .59

To get this outcome three times with replacement, we can do the following:

.59 \times .59 \times .59

\therefore .59^3

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 42%.

Answer: \frac{1}{1} \cdot \frac{49}{50} \cdot \frac{48}{50} \cdot \frac{47}{50}

We need to find the probability that there are no duplicates among the four teams selected for the tournament from bigkart with replacement. Since we are selecting four times, we want each selected team to be unique.

1. First Selection:
   • We can select any of the 50 teams, so the probability is \frac{50}{50} = 1.
2. Second Selection:
   • To avoid duplicates, we must pick a team different from the first one.
   • The probability is \frac{49}{50}.
3. Third Selection:
   • We must pick a team different from the first two selections.
   • The probability is \frac{48}{50}.
4. Fourth Selection:
   • We must pick a team different from the first three selections.
   • The probability is \frac{47}{50}.

The total probability that there are no duplicates among the four teams selected is the product of these probabilities: \frac{1}{1} \cdot \frac{49}{50} \cdot \frac{48}{50} \cdot \frac{47}{50}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: 1 - \frac{30}{50} \cdot \frac{29}{49} \cdot \frac{28}{48} \cdot \frac{27}{47}

We are selecting four teams from bigkart without replacement, and we want to calculate the probability that at least one Division 2 team is selected, which represented as P(A). We know that there are 30 Division 1 teams and 20 Division 2 teams.

First calculating the complement probability, P(Ac) which is the probability that all four teams are from Division 1.

1. First Selection:
   • Probability of picking a Division 1 team is \frac{30}{50}.
2. Second Selection:
   • After one Division 1 team is selected, there are 29 Division 1 teams left and 49 teams remaining in total.
   • The probability is \frac{29}{49}.
3. Third Selection:
   • After two Division 1 teams are selected, there are 28 Division 1 teams left and 48 teams remaining in total.
   • The probability is \frac{28}{48}.
4. Fourth Selection:
   • After three Division 1 teams are selected, there are 27 Division 1 teams left and 47 teams remaining in total.
   • The probability is \frac{27}{47}.

The probability that all four teams are from Division 1 is: \frac{30}{50} \cdot \frac{29}{49} \cdot \frac{28}{48} \cdot \frac{27}{47}

To find the probability of at least one Division 2 team being selected, we use P(A) = 1 - P(Ac): 1 - \frac{30}{50} \cdot \frac{29}{49} \cdot \frac{28}{48} \cdot \frac{27}{47}

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: False

The probability of rolling two fives can be found with 1/6 * 1/6 = 1/36. The probability of rolling a six and a three can be found with 2/6 (can roll either a 3 or 6) * 1/6 (roll a different side form 3 or 6, depending on what you rolled first) = 1/18. Therefore, the probabilities are not the same.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 33%.

Answer: 0.001

There is a \frac{1}{10} chance that we get 0 as the first random number, a \frac{1}{10} chance that we get 2 as the second random number, and a \frac{1}{10} chance that we get 7 as the third random number. The probability of all of these events happening is \frac{1}{10}*\frac{1}{10}*\frac{1}{10} = 0.001.

Another way to do this problem is to think about the possible outcomes. Any number from 000 to 999 is possible and all are equally likely. Since there are 1000 possible outcomes and the number 027 is just one of the possible outcomes, the probability of getting this outcome is \frac{1}{1000} = 0.001.

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Difficulty: ⭐️

The average score on this problem was 92%.

Answer: 0.5

Because the values of the left and right positions are not important to us, think of the middle position only. When selecting a random number to go here, we are choosing randomly from the numbers 0 through 9. Since 5 of these numbers are odd (1, 3, 5, 7, 9), the probability of getting an odd number is \frac{5}{10} = 0.5.

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Difficulty: ⭐️⭐️

The average score on this problem was 78%.

Answer: 0.271

It’s easier to calculate the probability that the number has no 7 in it, and then subtract this probability from 1. To solve this problem directly, we’d have to consider cases where 7 appeared multiple times, which would be more complicated.

The probability that the resulting number has no 7 is \frac{9}{10}*\frac{9}{10}*\frac{9}{10} = 0.729 because in each of the three positions, there is a \frac{9}{10} chance of selecting something other than a 7. Therefore, the probability that the number has a 7 is 1 - 0.729 = 0.271.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 69%.

Answer: \frac{1}{2^{31}}

First, let’s understand what df_choice does. It takes in one input, df. The line np.random.choice([True, False], df.shape[0], replace=True) evaluates to an array such that:

• There is one element for every row in the input DataFrame or array df (so if df has 31 rows, the output array will have length 31)
• Each element is equally likely to be True or False, since the sequence we are selecting from is [True, False] and we are selecting with replacement

So np.random.choice([True, False], df.shape[0], replace=True) is an array the same length as df, with each element randomly set to True or False. Note that there is a \frac{1}{2} chance the first element is True, a \frac{1}{2} chance the second element is True, and so on.

Then, df[np.random.choice([True, False], df.shape[0], replace=True)] is using Boolean indexing to keep only the rows in df where the array np.random.choice([True, False], df.shape[0], replace=True) contains the value True. So, the function df_choice returns a DataFrame containing somewhere between 0 and df.shape[0] rows. Note that there is a \frac{1}{2} chance that the new DataFrame contains the first row from df, a \frac{1}{2} chance that the new DataFrame contains the second row from df, and so on.

In this question, the only input ever given to df_choice is plum, which has 31 rows.

In this subpart, we’re asked for the probability that df_choice(plum) is an empty DataFrame. There are 31 rows, and each of them have a \frac{1}{2} chance of being included in the output, and so a \frac{1}{2} chance of being missing. So, the chance that they are all missing is:

\begin{aligned} P(\text{empty DataFrame}) &= P(\text{all rows missing}) \\ &= P(\text{row 0 missing and row 1 missing and ... and row 30 missing}) \\ &= P(\text{row 0 missing}) \cdot P(\text{row 1 missing}) \cdot ... \cdot P(\text{row 30 missing}) \\ &= \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot \frac{1}{2} \\ &= \boxed{\frac{1}{2^{31}}} \end{aligned}

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Difficulty: ⭐️⭐️

The average score on this problem was 83%.

Answer: \frac{31}{2^{31}}

In order for the resulting DataFrame to have 30 rows, exactly 1 row must be missing, and the other 30 must be present.

To start, let’s consider one row in particular, say, row 7. The probability that row 7 is missing is \frac{1}{2}, and the probability that rows 0 through 6 and 8 through 30 are all present is \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot \frac{1}{2} = \frac{1}{2^{30}} using the logic from the previous subpart. So, the probability that row 7 is missing AND all other rows are present is \frac{1}{2} \cdot \frac{1}{2^{30}} = \frac{1}{2^{31}}.

Then, in order for there to be 30 rows, either row 0 must be missing, or row 1 must be missing, and so on:

\begin{aligned} P(\text{exactly one row missing}) &= P(\text{only row 0 is missing or only row 1 is missing or ... or only row 30 is missing}) \\ &= P(\text{only row 0 is missing}) + P(\text{only row 1 is missing}) + ... + P(\text{only row 30 is missing}) \\ &= \frac{1}{2^{31}} + \frac{1}{2^{31}} + ... + \frac{1}{2^{31}} \\ &= \boxed{\frac{31}{2^{31}}} \end{aligned}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 48%.

Answer: True

An important realization to make is that all subsets of the rows in plum are equally likely to be returned by df_choice(plum), and they all have probability \frac{1}{2^{31}}. For instance, one subset of plum is the subset where rows 2, 5, 8, and 30 are missing, and the rest are all present. The probability that this subset is returned by df_choice(plum) is \frac{1}{2^{31}}.

This is true because for each individual row, the probability that it is present or missing is the same – \frac{1}{2} – so the probability of any subset is a product of 31 \frac{1}{2}s, which is \frac{1}{2^{31}}. (The answer to the previous subpart was not \frac{1}{2^{31}} because it was asking about multiple subsets – the subset where only row 0 was missing, and the subset where only row 1 was missing, and so on).

So, the probability that df_choice(plum) consists of just row 0 is \frac{1}{2^{31}}, and this is the same as the answer to the first subpart (\frac{1}{2^{31}}); in both situations, we are calculating the probability of one specific subset.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 63%.

Answer: \frac{1}{2^{25}}

Here, we’re not being asked for the probability of one specific subset (like the subset containing just row 0); rather, we’re being asked for the probability of various different subsets, so our calculation will be a bit more involved.

We can break our problem down into four pieces. We can find the probability that there are 0 rows, 1 row, 30 rows, and 31 rows individually, and add these probabilities up, since only one of them can happen at a time (it’s impossible for a DataFrame to have both 1 and 30 rows at the same time; these events are “mutually exclusive”). It turns out we’ve already calculated two of these probabilities:

• From the first subpart, the probability that 0 rows are returned is \frac{1}{2^{31}}. This corresponds to a single subset, the subset where all rows are missing.
• From the second subpart, the probability that 30 rows are returned is \frac{31}{2^{31}}.

The other two possibilities are symmetric with the above two!

• The probability that 31 rows are returned is the same as the probability that 0 rows are returned, since the probability of a row being missing and a row being present is the same. This is \frac{1}{2^{31}}.
• The probability that 1 row is returned is the same as the probability that 1 row is missing, i.e. the probability that 30 rows are returned. This, from the second subpart, is \frac{31}{2^{31}}.

Putting it all together, we have:

\begin{aligned} P(\text{number of returned rows is 0, 1, 30, or 31}) &= P(\text{0 rows are returned}) + P(\text{1 row is returned}) + P(\text{30 rows are returned}) + P(\text{31 rows are returned}) \\ &= \frac{1}{2^{31}} + \frac{31}{2^{31}} + \frac{31}{2^{31}} + \frac{1}{2^{31}} \\ &= \frac{1 + 31 + 31 + 1}{2^{31}} \\ &= \frac{64}{2^{31}} \\ &= \frac{2^6}{2^{31}} \\ &= \frac{1}{2^{31 - 6}} \\ &= \boxed{\frac{1}{2^{25}}} \end{aligned}

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 35%.

Answer: \frac{1}{2^{31}}

First, let’s understand what df_choice does. It takes in one input, df. The line np.random.choice([True, False], df.shape[0], replace=True) evaluates to an array such that:

• There is one element for every row in the input DataFrame or array df (so if df has 31 rows, the output array will have length 31)
• Each element is equally likely to be True or False, since the sequence we are selecting from is [True, False] and we are selecting with replacement

So np.random.choice([True, False], df.shape[0], replace=True) is an array the same length as df, with each element randomly set to True or False. Note that there is a \frac{1}{2} chance the first element is True, a \frac{1}{2} chance the second element is True, and so on.

Then, df[np.random.choice([True, False], df.shape[0], replace=True)] is using Boolean indexing to keep only the rows in df where the array np.random.choice([True, False], df.shape[0], replace=True) contains the value True. So, the function df_choice returns a DataFrame containing somewhere between 0 and df.shape[0] rows. Note that there is a \frac{1}{2} chance that the new DataFrame contains the first row from df, a \frac{1}{2} chance that the new DataFrame contains the second row from df, and so on.

In this question, the only input ever given to df_choice is plum, which has 31 rows.

In this subpart, we’re asked for the probability that df_choice(plum) is an empty DataFrame. There are 31 rows, and each of them have a \frac{1}{2} chance of being included in the output, and so a \frac{1}{2} chance of being missing. So, the chance that they are all missing is:

\begin{aligned} P(\text{empty DataFrame}) &= P(\text{all rows missing}) \\ &= P(\text{row 0 missing and row 1 missing and ... and row 30 missing}) \\ &= P(\text{row 0 missing}) \cdot P(\text{row 1 missing}) \cdot ... \cdot P(\text{row 30 missing}) \\ &= \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot \frac{1}{2} \\ &= \boxed{\frac{1}{2^{31}}} \end{aligned}

Answer: \frac{31}{2^{31}}

In order for the resulting DataFrame to have 30 rows, exactly 1 row must be missing, and the other 30 must be present.

To start, let’s consider one row in particular, say, row 7. The probability that row 7 is missing is \frac{1}{2}, and the probability that rows 0 through 6 and 8 through 30 are all present is \frac{1}{2} \cdot \frac{1}{2} \cdot ... \cdot \frac{1}{2} = \frac{1}{2^{30}} using the logic from the previous subpart. So, the probability that row 7 is missing AND all other rows are present is \frac{1}{2} \cdot \frac{1}{2^{30}} = \frac{1}{2^{31}}.

Then, in order for there to be 30 rows, either row 0 must be missing, or row 1 must be missing, and so on:

\begin{aligned} P(\text{exactly one row missing}) &= P(\text{only row 0 is missing or only row 1 is missing or ... or only row 30 is missing}) \\ &= P(\text{only row 0 is missing}) + P(\text{only row 1 is missing}) + ... + P(\text{only row 30 is missing}) \\ &= \frac{1}{2^{31}} + \frac{1}{2^{31}} + ... + \frac{1}{2^{31}} \\ &= \boxed{\frac{31}{2^{31}}} \end{aligned}

Answer: True

An important realization to make is that all subsets of the rows in plum are equally likely to be returned by df_choice(plum), and they all have probability \frac{1}{2^{31}}. For instance, one subset of plum is the subset where rows 2, 5, 8, and 30 are missing, and the rest are all present. The probability that this subset is returned by df_choice(plum) is \frac{1}{2^{31}}.

This is true because for each individual row, the probability that it is present or missing is the same – \frac{1}{2} – so the probability of any subset is a product of 31 \frac{1}{2}s, which is \frac{1}{2^{31}}. (The answer to the previous subpart was not \frac{1}{2^{31}} because it was asking about multiple subsets – the subset where only row 0 was missing, and the subset where only row 1 was missing, and so on).

So, the probability that df_choice(plum) consists of just row 0 is \frac{1}{2^{31}}, and this is the same as the answer to the first subpart (\frac{1}{2^{31}}); in both situations, we are calculating the probability of one specific subset.

Answer: \frac{1}{2^{25}}

Here, we’re not being asked for the probability of one specific subset (like the subset containing just row 0); rather, we’re being asked for the probability of various different subsets, so our calculation will be a bit more involved.

We can break our problem down into four pieces. We can find the probability that there are 0 rows, 1 row, 30 rows, and 31 rows individually, and add these probabilities up, since only one of them can happen at a time (it’s impossible for a DataFrame to have both 1 and 30 rows at the same time; these events are “mutually exclusive”). It turns out we’ve already calculated two of these probabilities:

• From the first subpart, the probability that 0 rows are returned is \frac{1}{2^{31}}. This corresponds to a single subset, the subset where all rows are missing.
• From the second subpart, the probability that 30 rows are returned is \frac{31}{2^{31}}.

The other two possibilities are symmetric with the above two!

• The probability that 31 rows are returned is the same as the probability that 0 rows are returned, since the probability of a row being missing and a row being present is the same. This is \frac{1}{2^{31}}.
• The probability that 1 row is returned is the same as the probability that 1 row is missing, i.e. the probability that 30 rows are returned. This, from the second subpart, is \frac{31}{2^{31}}.

Putting it all together, we have:

\begin{aligned} P(\text{number of returned rows is 0, 1, 30, or 31}) &= P(\text{0 rows are returned}) + P(\text{1 row is returned}) + P(\text{30 rows are returned}) + P(\text{31 rows are returned}) \\ &= \frac{1}{2^{31}} + \frac{31}{2^{31}} + \frac{31}{2^{31}} + \frac{1}{2^{31}} \\ &= \frac{1 + 31 + 31 + 1}{2^{31}} \\ &= \frac{64}{2^{31}} \\ &= \frac{2^6}{2^{31}} \\ &= \frac{1}{2^{31 - 6}} \\ &= \boxed{\frac{1}{2^{25}}} \end{aligned}

Answer: 0.025

In order for the 8th floor of Building C to be selected, two things need to happen:

• In Wave 1, the 8th floor of Building C needs to be selected amongst all 10 floors in Building C – this happens with probability \frac{1}{10}, since each floor in Building C is equally likely to be selected.
• In Wave 2, the 8th floor of Building C needs to be selected amongst the 4 floors selected in Wave 1 – this happens with probability \frac{1}{4}, since each floor in Wave 1 is equally likely to be selected.

Since both events need to occur, and both events are independent (think of selecting in each wave as drawing names from a hat), the probability that both occur is the product of the probabilities that they occur individually:

\frac{1}{10} \cdot \frac{1}{4} = \frac{1}{40} = 0.025

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: 0.344

Whenever we are asked to compute the probability of “at least one” occurrence of some event, it is almost always easiest to compute the complement of (i.e. “1 minus”) the probability that there are no occurrences of the event. That is the case here; as such, we need to compute the probability that none of the 10th floors are selected across all four buildings.

To compute the probability that none of the 10th floors are selected across all four buildings, we first need to find the probability that the 10th floor is not selected in just a single building. This is 1 - \frac{1}{10} = \frac{9}{10}. Then, since the selections in each building are independent of other buildings, the probability that none of the 10th floors are selected across all four buildings is \left( \frac{9}{10} \right)^4.

Lastly, the probability we are asked for is the complement of the probability we just computed. So, the probability that at least one 10th floor is selected across all four buildings is

1 - \left( 1 - \frac{1}{10} \right)^4 = 1 - \left( \frac{9}{10} \right)^4 = 1 - 0.6561 = 0.3439

This rounds to 0.344.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: Series

strategy will give you a Series. This is because games.get(“Domains”) will give you one column, a Series, and then .str.contains(“Strategy Games”) will convert those values to True if it contains that string and False otherwise, but it will not actually change the Series to a DataFrame, a bool, or a str.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: (games[(strategy == True) & (thematic == False)].shape[0] / games[strategy == True].shape[0]) or 1 - games[strategy & thematic].shape[0] / games[strategy].shape[0]

The problem is asking us to find the probability that a selected game from “Strategy Games” will not be in “Thematic Games”. Recall that this is the probability that given “Strategy Games” will it not be in “Thematic Games”, which would look like this: P(“Thematic Games” Compliment|“Strategy Games”). This means the formula would look like: (“Thematic Games” Compliment and “Strategy Games”)/“Strategy Games”

This means one possible solution for this would be: (games[(strategy == True) & (thematic == False)].shape[0] / games[strategy == True].shape[0] This solution works because we are following the formula to find the probability of thematic complement and strategy games over the number of times “Strategy Games” are True. Doing games[query_condition] gives us the games DataFrame where strategy == True and thematic == False. Another important thing is that for (baby)pandas we always use the keyword & and not and. Note that we are using .shape[0] to get the number of rows or times that True shows up for “Strategy Games” and the number of rows or times that False shows up for “Thematic Games” Compliment.

Another possible strategy would be using the complement rule: Which would be: P(“Thematic Games” Compliment) = 1 - P(“Thematic Games”). This would lead you to an answer like: (1 - games[strategy & thematic].shape[0]) / games[strategy].shape[0]. games[strategy & thematic].shape[0]) finds you the probability of P(“Thematic Games”), so when plugged into the equation above we are able to find P(“Thematic Games” Compliment).

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 43%.

Answer: Options 3 and 4

Let’s take a closer look at why Option 3 and Option 4 are correct.

Option 3: Option 3 first queries the games DataFrame to only keep games with one “Domains”. games.get(“Domains”).str.split(“,”).apply(len) == 1 gets the “Domains” column and splits all of them if they contain a comma. If the value does have a comma then it will create a list. For example let’s say the domain was “Strategy Games”, “Thematic Games” then after doing str.split(“,”) we would have the list: [“Strategy Games”, “Thematic Games”]. Any row with a list will evaluate to False. This means we are only keeping values where there is one domain. The next part .groupby(“Domains”).count().get(“Name”).sum() makes a DataFrame with an index of the unique domains and the number of times those appear. Note that all the other columns: “Name”, “Mechanics”, “Play Time”, “Complexity”, “Rating”, and “BGG Rank” now evaluate to the same thing, the number of times a unique domain appears. That means by doing .get(“Name”).sum() we are adding up all the number of times a unique domain appears, which would get us the number of games that belong to only one domain.

Option 4: Option 4 starts off exactly like Option 3, but instead of doing .groupby() it gets the number of rows using .shape[0], which will give us the number of games that belong to only one domain.

Option 1: Let’s step through why Option 1 is incorrect. games.get(“Domains”).str.split(“ ”).apply(len) == 2.sum() gives you a Series of the “Domains” column, then splits each domain by a space. We then get the length of that list, check if the length is equal to 2, which would mean there are two elements in the list, and finally get the sum of all elements in the list who had two elements because of the split. Remember that True evaluates to 1 and False evaluates to 0, so we are getting the number of elements that were split into two. It does not tell us the number of games that belong to only one domain.

Option 2: Let’s step through why Option 2 is also incorrect. (games.get(“Domains”).apply(len) == 1).sum() checks to see if each element in the column “Domains” has only one character. Remember when you apply len() to a string then we get the number of characters in that string. This is essentially counting the number of domains that have 1 letter. Thus, it does not tell us the number of games that belong to only one domain.

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: np.count_nonzero(landfalls == "Georgia") / 10000

The probability would be the number of times the storm hit Georgia over the total number of simulated hurricanes, which is 10000. Since each element of landfills is a string, we can use np.count_nonzero() to give us all of the times the storm hits Georgia in the simulation. Recall np.count_nonzero() determines whether the elements are “truthy”, so if the String was equal to "Georgia" then it would be counted, and otherwise would be ignored. Then to calculate the probability we simply divide the number of times the storm hit Georgia by 10000.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 34%.

Answer:

Originally, we were unsure if the hurricane would make landfall. Now that it has our total percentage has changed: 100\% - 25\% = 75\%. Remember probability should always add up to 100%. So our new total is 75%. We can calculate the probability that a storm hits Georgia by doing: \frac{25\%}{75\%} = \frac{1}{3}.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 36%.

Answer: \frac{1}{4}

Reading the prompt we can see that we are solving for a conditional probability. Let A be the given condition that the medal is gold and let B be the event that a medal is from 2000. Looking at the DataFrame we can see that 8 total gold medals are earned (make sure you pay attention to the count column). Out of these 8 medals, 2 of them are from the year 2000. Thus, we obtain the probability \frac{2}{8} or \frac{1}{4}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 70%.

Answer: \frac{3}{4}

Here we can recognize that we are solving for the probability of a union of two events. Let A be the event that the medal is gold. Let B be the event that it is earned while representing East Germany. The probability formula for a union is P(A \cup B) = P(A) + P(B) - P(A \cap B). Looking at the DataFrame, we know P(A)=\frac{8}{12}, P(B)=\frac{4}{12}, and P(A \cap B)=\frac{3}{12}. Plugging all of this into the formula, we get \frac{8}{12}+\frac{4}{12}-\frac{3}{12}=\frac{9}{12}=\frac{3}{4}. Thus, the correct answer is \frac{3}{4}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 73%.

Answer: \frac{1}{22}

In this problem, we are sampling 2 medals without replacement. Let A be the event that the first medal is from 1988 and let P(B) be the event that the second medal is from 1988. P(A) is \frac{3}{12} since there are 3 medals from 1988 out of the total 12 medals. However, in the second trial, since we are sampling without replacement, the medal that we just sampled is no longer in our pool. Thus, P(B) is now \frac{2}{11} since there are now 2 medals from 1988 from the remaining 11 total medals. The joint probability P(A \cap B) is P(A)P(B) so, plugging these values in, we get \frac{3}{12} \cdot \frac{2}{11} = \frac{3}{66} = \frac{1}{22}. Thus, the answer is \frac{1}{22}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 57%.

Answer: \frac{6}{19}

The probability of getting a weapon card is just the number of weapon cards divided by the total number of cards. There are 6 weapon cards and 19 cards total, so the probability has to be \frac{6}{19}. Note that it does not matter how the cards were dealt. Though each card is dealt one at a time to each player, Janine will always end up with a randomly selected 6 cards, out of the 19 cards available.

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Difficulty: ⭐️⭐️

The average score on this problem was 80%.

Answer: \frac{6}{19} \cdot \frac{5}{18} \cdot \frac{4}{17} \cdot \frac{3}{16} \cdot \frac{2}{15} \cdot \frac{1}{14}

We can calculate the answer using the multiplication rule. The probability of getting Janine getting all the weapon cards is the probability of getting a dealt a weapon card first multiplied by the probability of getting a weapon card second multiplied by continuing probabilities of getting a weapon card until probability of getting a weapon card on the sixth draw. The denominator of each subsequent probability decreases by 1 because we remove one card from the total number of cards on each draw. The numerator also decreases by 1 because we remove a weapon card from the total number of available weapon cards on each draw.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 62%.

Answer: k = 12, m = 18

m has to be 18 because the denominator is the number of cards available during the first and second draw. We have 19 cards on the first draw and 18 on the second draw, so the only way to get that is for m = 18.

The probability that exactly one of the cards of your first two draws is a weapon card can be broken down into two cases: getting a weapon card first and then a non-weapon card, or getting a non-weapon card first and then a weapon card. We add the probabilities of the two cases together in order to calculate the overall probability, since the cases are mutually exclusive, meaning they cannot both happen at the same time.

Consider first the probability of getting a weapon card followed by a non-weapon card. This probability is \frac{6}{19} \cdot \frac{13}{18}. Similarly, the probability of getting a non-weapon card first, then a weapon card, is \frac{13}{19} \cdot \frac{6}{18}. The sum of these is \frac{6 \cdot 13}{19 \cdot 18} + \frac{13 \cdot 6}{19 \cdot 18}.

Since we want the numerator to look like k \cdot (k+1), we want to combine the terms in the numerator. Since the fractions in the sum are the same, we can represent the probability as 2 \cdot \frac{6}{19} \cdot \frac{13}{18}. Since 2\cdot 6 = 12, we can express the numerator as 12 \cdot 13, so k = 12.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 31%.

Answer: \left(0.25\right)^{7}

The probability that a student gets assigned to Gryffindor is 0.25 (or \frac{1}{4}). Since we are given that the Sorting Hat assigns students independently, we can apply the multiplication rule and multiply the probability of each of the seven Weasleys being sorted to Gryffindor together. Therefore, all terms in our product are 0.25. Thus, the probability that all seven Weasleys get assigned to Gryffindor would be 0.25 * 0.25 * \ldots * 0.25 = \left(0.25\right)^{7}.

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: 0.25

Fred and George are assigned to the same house if both are in Gryffindor, or both are in Hufflepuff, etc.. Fred and George’s selections from the sorting hat are independent. Using the multiplication rule, P(both in Gryffindor) = 0.25 * 0.25. Likewise, P(both in Hufflepuff) = 0.25 * 0.25. It is the same for Ravenclaw and Slytherin as well. Therefore, P(same house) = 4 * (0.25 * 0.25) = 0.25.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

Answer: \left(0.75\right)^{7}

The probability that none of the seven Weasley siblings are assigned to Slytherin is equal to the probability that all of the Weasley siblings are not assigned to Slytherin. The probability that someone is not assigned to Slytherin is 1 - P(Slytherin) = 1 - 0.25 = 0.75. Since we are given that the Sorting Hat assigns students independently, we can apply the multiplication rule and multiply the probability of each of the seven Weasleys being sorted to anywhere but Slytherin together. Therefore, all terms in our product are 0.75. Thus, the probability that none of seven Weasleys get assigned to Slytherin would be 0.75 * 0.75 * \ldots * 0.75 = \left(0.75\right)^{7}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: 1 - \left(\frac{2}{3}\right)^{7}

Since we are given that none of the Weasley siblings are assigned to Slytherin, our possibilities are now only Gryffindor, Hufflepuff, and Ravenclaw. Therefore, the probability that someone is assigned to Gryffindor would become \frac{1}{3}. Correspondingly, the probability that someone isn’t assigned to Gryffindor is 1 - \frac{1}{3} = \frac{2}{3}. The probability that at least one of the siblings is assigned to Gryffindor is equal to 1 - the probability that none of the siblings are assigned to Gryffindor, or 1 - \left(\frac{2}{3}\right)^{7}.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 51%.