Answer: -4

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: 35

Let’s attempt this problem step-by-step. We’ll first determine the number of rows in tesla.merge(bmw, on="BodyStyle"), and then determine the number of rows in combo. For the purposes of the solution, let’s use temp to refer to the first merged DataFrame, tesla.merge(bmw, on="BodyStyle").

Recall, when we merge two DataFrames, the resulting DataFrame contains a single row for every match between the two columns, and rows in either DataFrame without a match disappear. In this problem, the column that we’re looking for matches in is "BodyStyle".

To determine the number of rows of temp, we need to determine which rows of tesla have a "BodyStyle" that matches a row in bmw. From the DataFrame provided, we can see that the only "BodyStyle"s in both tesla and bmw are SUV and sedan. When we merge tesla and bmw on "BodyStyle":

• The 4 SUV rows in tesla each match the 1 SUV row in bmw. This will create 4 SUV rows in temp.
• The 3 sedan rows in tesla each match the 1 sedan row in bmw. This will create 3 sedan rows in temp.

So, temp is a DataFrame with a total of 7 rows, with 4 rows for SUVs and 3 rows for sedans (in the "BodyStyle") column. Now, when we merge temp and audi on "BodyStyle":

• The 4 SUV rows in temp each match the 8 SUV rows in audi. This will create 4 \cdot 8 = 32 SUV rows in combo.
• The 3 sedan rows in temp each match the 1 sedan row in audi. This will create 3 \cdot 1 = 3 sedan rows in combo.

Thus, the total number of rows in combo is 32 + 3 = 35.

Note: You may notice that 35 is the result of multiplying the "SUV" and "Sedan" columns in the DataFrame provided, and adding up the results. This problem is similar to Problem 5 from the Fall 2021 Midterm.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 45%.

Answer: np.array([]) or []

res is the list in which we’ll be storing each cumulative sum. Thus we start by initializing res to an empty array or list.

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Difficulty: ⭐️

The average score on this problem was 100%.

Answer: range(1, len(arr)) or np.arange(1, len(arr))

We’re trying to loop through the indices of arr and calculate the cumulative sum corresponding to each entry. To access each index in sequential order, we simply use range() or np.arange(). However, notice that we have already appended the first entry of arr to res on line 3 of the code snippet. (Note that the first entry of arr is the same as the first cumulative sum.) Thus the lower bound of range() (or np.arange()) actually starts at 1, not 0. The upper bound is still len(arr) as usual.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 64%.

Answer: res[i - 1] + arr[i] or sum(arr[:i + 1])

Looking at the syntax of the problem, the blank we have to fill essentially requires us to calculate the current cumulative sum, since the rest of line will already append the blank to res for us. One way to think of a cumulative sum is to add the “current” arr element to the previous cumulative sum, since the previous cumulative sum encapsulates all the previous elements. Because we have access to both of those values, we can easily represent it as res[i - 1] + arr[i]. The second answer is more a more direct approach. Because the cumulative sum is just the sum of all the previous elements up to the current element, we can directly compute it with sum(arr[:i + 1])

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

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: 108

Since we used the argument on="flips, rows from teresa and sophia will be combined whenever they have matching values in their "flips" columns.

For the teresa DataFrame:

• There are 13 rows with "Heads" in the "flips" column.
• There are 8 rows with "Tails" in the "flips" column.

For the sophia DataFrame:

• There are 4 rows with "Heads" in the "flips" column.
• There are 7 rows with "Tails" in the "flips" column.

The merged DataFrame will also only have the values "Heads" and "Tails" in its "flips" column. - The 13 "Heads" rows from teresa will each pair with the 4 "Heads" rows from sophia. This results in 13 \cdot 4 = 52 rows with "Heads" - The 8 "Tails" rows from teresa will each pair with the 7 "Tails" rows from sophia. This results in 8 \cdot 7 = 56 rows with "Tails".

Then, the total number of rows in the merged DataFrame is 52 + 56 = 108.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 54%.

Answer: A+1

The additional row in each DataFrame has a unique "flips" value of "Total". When we merge on the "flips" column, this unique value will only create a single new row in the merged DataFrame, as it pairs the "Total" from teresa with the "Total" from sophia. The rest of the rows are the same as in the previous merge, and as such, they will contribute the same number of rows, A, to the merged DataFrame. Thus, the total number of rows in the new merged DataFrame will be A (from the original matching rows) plus 1 (from the new "Total" rows), which sums up to A+1.

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Difficulty: ⭐️⭐️⭐️⭐️

The average score on this problem was 46%.

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: 6 8

Imagine there is a pointer from a to 6 and from b to 4. In the function we look at the first if-statement. The if-statement is checking if a, which points to 6, is less than b, which points to 4. We know 6 is not less than 4, so we skip this section of code. Next we see the second if-statement which checks if a, 6, plus 2 equals b, 4. We know 6 + 2 = 8, which is not equal to 4. We then look at the elif-statement which asks if a, 6, minus 1 is greater than b, 4. This is True! 6 - 1 = 5 and 5 > 4. So we print(a), which will spit out 6 and then we will return a + 2. a + 2 is 6 + 2. This means the function wham will print 6 and return 8.

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Difficulty: ⭐️⭐️

The average score on this problem was 81%.

Answer: Any pair of integers a, b with a = b or with a = b + 1

The desired output is a + 1. So we want to look at the function wham and see which condition is necessary to get the output a + 1. It turns out that this can be found in the else-block, which means we need to find an a and b that will not satisfy any of the if or elif-statements.

If a = b, so for example a points to 4 and b points to 4 then: a is not less than b (4 < 4), a + 2 is not equal to b (4 + 2 = 6 and 6 does not equal 4), and a - 1 is not greater than b (4 - 1= 3) and 3 is not greater than 4.

If a = b + 1 this means that a is greater than b, so for example if b is 4 then a is 5 (4 + 1 = 5). If we look at the if-statements then a < b is not true (5 is greater than 4), a + 2 == b is also not true (5 + 2 = 7 and 7 does not equal 4), and a - 1 > b is also not true (5 - 1 = 4 and 4 is equal not greater than 4). This means it will trigger the else statement.

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: 6

For this to happen: a + 2 == b then a must be less than b by 2. However if a is less than b it will trigger the first if-statement. This means this second if-statement will never run, which means that the return on line 6 never happens.

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

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: ['Year', 'Appearance_Order']

The fact that groupby automatically sorts the index in ascending order is important here. Since we want earlier years before later years, we could group by 'Year', however if we just group by year, all the artists who performed in a given year will be aggregated together, which is not what we want. Within each year, we want to organize the artists in ascending order of 'Appearance_Order'. In other words, we need to group by 'Year' with 'Appearance_Order' as subgroups. Therefore, the correct way to reorder the rows of sungod as desired is sungod.groupby(['Year', 'Appearance_Order']).max().reset_index(). Note that we need to reset the index so that the resulting DataFrame has 'Year' and 'Appearance_Order' as columns, like in sungod.

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Difficulty: ⭐️⭐️

The average score on this problem was 85%.

Answer: merged = sungod.merge(music, left_on='Artist', right_on='Name')

The question we want to answer is about Sun God music artists’ genres. In order to answer, we’ll need a DataFrame consisting of rows of artists that have performed at Sun God since its inception in 1983. If we merge the sungod DataFrame with the music DataFrame based on the artist’s name, we’ll end up with a DataFrame containing one row for each artist that has ever performed at Sun God. Since the column containing artists’ names is called 'Artist' in sungod and 'Name' in music, the correct syntax for this merge is merged = sungod.merge(music, left_on='Artist', right_on='Name'). Note that we could also interchange the left DataFrame with the right DataFrame, as swapping the roles of the two DataFrames in a merge only changes the ordering of rows and columns in the output, not the data itself. This can be written in code as merged = music.merge(sungod, left_on='Name', right_on='Artist'), but this is not one of the answer choices.

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Difficulty: ⭐️⭐️

The average score on this problem was 86%.

Answer: top_hit_year - year

Before we can answer this question, we need to understand what the first three lines of the classify_artist function are doing. The first line creates a DataFrame with only one row, corresponding to the particular artist that’s passed in as input to the function. We know there is just one row because we are told that the artist being passed in as input has appeared exactly once at Sun God. The next two lines create two variables:

• year contains the year in which the artist performed at Sun God, and
• top_hit_year contains the year in which their top hit song was released.

Now, we can fill in blank (a). Notice that the body of the if clause is return 'up-and-coming'. Therefore we need a condition that corresponds to up-and-coming, which we are told means the top hit came out after the artist appeared at Sun God. Using the variables that have been defined for us, this condition is top_hit_year > year. However, the if statement condition is already partially set up with > 0 included. We can simply rearrange our condition top_hit_year > year by subtracting year from both sides to obtain top_hit_year - year > 0, which fits the desired format.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

Answer: year-top_hit_year > 5

For this part, we need a condition that corresponds to an artist being outdated which happens when their top hit came out more than five years prior to their appearance at Sun God. There are several ways to state this condition: year-top_hit_year > 5, year > top_hit_year + 5, or any equivalent condition would be considered correct.

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Difficulty: ⭐️⭐️

The average score on this problem was 89%.

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%.