Answer: takes a string as input, returns a string

To use the Series method .apply, we first need a Series, containing values of any type. We pass in the name of a function to .apply and essentially, .apply calls the given function on each value of the Series, producing a Series with the resulting outputs of those function calls. In this case, .apply(extract_product_line) is called on the Series ikea.get('product'), which contains string values. This means the function extract_product_line must take strings as inputs. We’re told that the code assigns a new column to the ikea DataFrame containing the product line associated with each product, and we know that the product line is a string, as it’s the first word of the product name. This means the function extract_product_line must output a string.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: x.split(' ')[0]

This function should take as input a string x, representing a product name, and return the first word of that string, representing the product line. Since words are separated by spaces, we want to split the string on the space character ' '.

It’s also correct to answer x.split()[0] without specifying to split on spaces, because the default behavior of the string .split method is to split on any whitespace, which includes any number of spaces, tabs, newlines, etc. Since we’re only extracting the first word, which will be separated from the rest of the product name by a single space, it’s equivalent to split using single spaces and using the default of any whitespace.

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Difficulty: ⭐️⭐️

The average score on this problem was 84%.

Answer: int(first_split[0])*60+int(second_split[0])

As the last subpart demonstrated, if we want to compare times, it doesn’t make sense to do so when times are represented as strings. In the to_minutes function, we convert a time string into an integer number of minutes.

The first step is to understand the logic. Every hour contains 60 minutes, so for a time string formatted like x hr, y min' the total number of minutes comes from multiplying the value of x by 60 and adding y.

The second step is to understand how to extract the x and y values from the time string using the string methods .split. The string method .split takes as input some separator string and breaks the string into pieces at each instance of the separator string. It then returns a list of all those pieces. The first line of code, therefore, creates a list called first_split containing two elements. The first element, accessed by first_split[0] contains the part of the time string that comes before ' hr, '. That is, first_split[0] evaluates to the string x.

Similarly, first_split[1] contains the part of the time string that comes after ' hr, '. So it is formatted like 'y min'. If we split this string again using the separator of ' min', the result will be a list whose first element is the string 'y'. This list is saved as second_split so second_split[0] evaluates to the string y.

Now we have the pieces we need to compute the number of minutes, using the idea of multiplying the value of x by 60 and adding y. We have to be careful with data types here, as the bolded instructions warn us that the function must return an integer. Right now, first_split[0] evaluates to the string x and second_split[0] evaluates to the string y. We need to convert these strings to integers before we can multiply and add. Once we convert using the int function, then we can multiply the number of hours by 60 and add the number of minutes. Therefore, the solution is int(first_split[0])*60+int(second_split[0]).

Note that failure to convert strings to integers using the int function would lead to very different behavior. Let’s take the example time string of '3 hr, 5 min' as input to our function. With the return statement as int(first_split[0])*60+int(second_split[0]), the function would return 185 on this input, as desired. With the return statement as first_split[0]*60+second_split[0], the function would return a string of length 61, looking something like this '3333...33335'. That’s because the * and + symbols do have meaning for strings, they’re just different meanings than when used with integers.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 71%.

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: 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: 6 8

When we call wham(6, 4), a gets assigned to the number 6 and b gets assigned to the number 4. In the function we look at the first if-statement. The if-statement is checking if a, 6, is less than b, 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: ['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: 38

There are two flights from LAX. When we merge socal with itself on the 'FROM' column, each of these flights gets paired up with each of these flights, for a total of four rows in the output. That is, the first flight from LAX gets paired with both the first and second flights from LAX. Similarly, the second flight from LAX gets paired with both the first and second flights from LAX.

Following this logic, each of the five flights from SAN gets paired with each of the five flights from SAN, for an additional 25 rows in the output. For SNA, there will be 9 rows in the output. The total is therefore 2^2 + 5^2 + 3^2 = 4 + 25 + 9 = 38 rows.

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Difficulty: ⭐️⭐️⭐️⭐️⭐️

The average score on this problem was 27%.

Answer: ['FROM', 'TO']

We want to organize flights by route. This means we need to group by both 'FROM' and 'TO' so any flights with the same pair of departure and arrival airports get grouped together. To group by multiple columns, we must use a list containing all these column names, as in flights.groupby(['FROM', 'TO']).

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 72%.

Answer: sum()

Every .groupby command needs an aggregation function! Since we are asked to find the route that King Triton has spent the most time flying on, we want to total the times for all flights on a given route.

Note that .count() would tell us how many flights King Triton has taken on each route. That’s meaningful information, but not what we need to address the question of which route he spent the most time flying on.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 58%.

Answer: by='HOURS', ascending=False

We want to know the route that King Triton spent the most time flying on. After we group flights by route, summing flights on the same route, the 'HOURS' column contains the total amount of time spent on each route. We need most_frequent.get('FROM').iloc[0] and most_frequent.get('TO').iloc[0] to correspond with the departure and destination airports of the route that King Triton has spent the most time flying on. To do this, we need to sort in descending order of time, to bring the largest time to the top of the DataFrame. So we must sort by 'HOURS' with ascending=False.

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Difficulty: ⭐️

The average score on this problem was 94%.

Answer: Option 1, Option 2, Option 3

All three options start with the same first line of code: month_as_num = int(date.split('-')[1]). This takes the date, originally a string formatted such as '2021-09-07', separates it into a list of three strings such as ['2021', '09', '07'], extracts the element in position 1 (the middle position), and converts it to an int such as 9. Now we have the month as a number we can work with more easily.

According to the definition of seasons, the months in each season are as follows:

Season	Month	month_as_num
Spring	March, April, May	3, 4, 5
Summer	June, July, August	6, 7, 8
Fall	September, October, November	9, 10, 11
Winter	December, January, February	12, 1, 2

Option 1 correctly assigns months to seasons by checking if the month falls in the appropriate range for 'Spring', then 'Summer', then 'Fall'. Finally, if all of these conditions are false, the else branch will return the correct answer of 'Winter' when month_as_num is 12, 1, or 2.

Option 2 is also correct, and in fact, it does the same exact thing as Option 1 even though it uses if where Option 1 used elif. The purpose of elif is to check a condition only when all previous conditions are false. So if we have an if followed by an elif, the elif condition will only be checked when the if condition is false. If we have two sequential if conditions, typically the second condition will be checked regardless of the outcome of the first condition, which means two if statements can behave differently than an if followed by an elif. In this case, however, since the if statements cause the function to return and therefore stop executing, the only way to get to a certain if condition is when all previous if conditions are false. If any prior if condition was true, the function would have returned already! So this means the three if conditions in Option 2 are equivalent to the if, elif, elif structure of Option 1. Note that the else case in Option 1 is reached when all prior conditions are false, whereas the else in Option 2 is paired only with the if statement immediately preceding it. But since we only ever get to that third if statement when the first two if conditions are false, we still only reach the else branch when all three if conditions are false.

Option 3 works similarly to Option 1, except it separates the months into more categories, first categorizing January and February as 'Winter', then checking for 'Spring', 'Summer', and 'Fall'. The only month that winds up in the else branch is December. We can think of Option 3 as the same as Option 1, except the Winter months have been separated into two groups, and the group containing January and February is extracted and checked first.

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Difficulty: ⭐️⭐️

The average score on this problem was 76%.

Answer: flights.get('DATE').apply(date_to_season)

Our function date_to_season takes as input a single date and converts it to a season. We cannot input a whole Series of dates, as in the first answer choice. We instead need to apply the function to the whole Series of dates. The correct syntax to do that is to first extract the Series of dates from the DataFrame and then use .apply, passing in the name of the function we wish to apply to each element of the Series. Therefore, the correct answer is flights.get('DATE').apply(date_to_season).

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Difficulty: ⭐️

The average score on this problem was 97%.

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: 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: cookie_box, 2, replace=False

We are told that King Triton grabs two cookies out of the box without looking. Since this is a random choice, we use the function np.random.choice to simulate this. The first input to this function is a sequence of values to choose from. We already have an array of values to choose from in the variable cookie_box. Calling np.random.choice(cookie_box) would select one cookie from the cookie box, but we want to select two, so we use an optional second parameter to specify the number of items to randomly select. Finally, we should consider whether we want to select with or without replacement. Since cookie_box contains individual cookies and King Triton is selecting two of them, he cannot choose the same exact cookie twice. This means we should sample without replacement, by specifying replace=False. Note that omitting the replace parameter would use the default option of sampling with replacement.

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Difficulty: ⭐️

The average score on this problem was 92%.

Answer: grab[0] == grab[1]

The idea of a simulation is to do some random process many times. We can use the results to approximate a probability by counting up the number of times some event occurred, and dividing that by the number of times we did the random process. Here, the random process is selecting two cookies from the cookie box, and we are doing this 10,000 times. The approximate probability will be the number of times in which both cookies are the same divided by 10,000. So we need to count up the number of times that both randomly selected cookies are the same. We do this by having an accumulator variable that starts out at 0 and gets incremented, or increased by 1, every time both cookies are the same. The code has such a variable, called prob_both_same, that is initialized to 0 and gets incremented when some condition is met.

We need to fill in the condition, which is that both randomly selected cookies are the same. We’ve already randomly selected the cookies and stored the results in grab, which is an array of length 2 that comes from the output of a call to np.random.choice. To check if both elements of the grab array are the same, we access the individual elements using brackets with the position number, and compare using the == symbol to check equality. Note that at the end of the for loop, the variable prob_both_same will contain a count of the number of trials out of 10,000 in which both of King Triton’s cookies were the same flavor.

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Difficulty: ⭐️⭐️

The average score on this problem was 79%.

Answer: prob_both_same / repetitions

After the for loop, prob_both_same contains the number of trials out of 10,000 in which both of King Triton’s cookies were the same flavor. We’d like it to represent the approximate probability of both cookies being the same flavor, so we need to divide the current value by the total number of trials, 10,000. Since this value is stored in the variable repetitions, we can divide prob_both_same by repetitions.

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Difficulty: ⭐️

The average score on this problem was 93%.

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: attending.sum() > 20000

Let’s look at the variable attending. Since we’re choosing 21,000 elements from the list [True, False] and there are 21,000 tickets distributed, this code is randomly determining whether each ticket holder will actually attend the festival. There’s a 95% chance of each ticket holder attending, which is reflected in the p=[0.95, 0.05] argument. Remember that np.random.choice returns an array of random choices, which in this case means it will contain 21,000 elements, each of which is True or False.

We want to figure out the probability of at least one ticket holder showing up and not being admitted. Another way to say this is we want to find the probability that more than 20,000 ticket holders show up to attend the festival. The way we approximate a probability through simulation is we repeat a process many times and see how often some event occured. The event we’re interested in this case is that more than 20,000 ticket holders came to Sun God. Since we have an array of True and False values corresponding to whether each ticket holder actually came, we just need to determine if there are more than 20,000 True values in the attending array.

There are several ways to count the number of True values in a Boolean array. One way is to sum the array since in Python True counts as 1 and False counts as 0. Therefore, attending.sum() > 20000 is the condition we need to check here.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 67%.

Answer: num_angry = num_angry + 1

Remember our goal in simulation is to repeat a process many times to see how often some event occurs. The repetition comes from the for loop which runs 10,000 times. Each time, we are simulating the process of 21,000 students each randomly deciding whether to show up to Sun God or not. We want to know, out of these 10,000 trials, how frequently more than 20,000 of the students will show up. So when this happens, we want to record that it happened. The standard way to do that is to keep a counter variable that starts at 0 and gets incremented, or increased by one, each time we had more than 20,000 attendees in our simulation.

The framework to do this is already set up because a variable called num_angry is initialized to 0 before the for loop. This variable is our counter variable, meant to count the number of trials, out of 10,000, that resulted in at least one student being angry because they showed up to Sun God with a ticket and were denied entrance. So all we need to do when there are more than 20,000 True values in the attending array is increment this counter by one via the code num_angry = num_angry + 1, sometimes abbreviated as num_angry += 1.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 59%.

Answer: num_angry/10000

To calculate the approximate probability, all we need to do is divide the number of trials in which a student was angry by the total number of trials, which is 10,000.

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Difficulty: ⭐️⭐️⭐️

The average score on this problem was 68%.

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: 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: 0.2

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: \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%.