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The problems in this worksheet are taken from past exams. Work on
them on paper, since the exams you take in this course
will also be on paper.
We encourage you to complete this
worksheet in a live discussion section. Solutions will be made available
after all discussion sections have concluded. You don’t need to submit
your answers anywhere.
Note: We do not plan to cover all
problems here in the live discussion section; the problems we don’t
cover can be used for extra practice.
Suppose you are booking a flight and you have no control over which airline you fly on. Below is a table with multiple airlines and the probability of a flight being on a specific airline.
| Airline | Chance |
|---|---|
| Delta | 0.4 |
| United | 0.3 |
| American | 0.2 |
| All other airlines | 0.1 |
The airline for one flight has no impact on the airline for another flight.
For this question, suppose that you schedule 3 flights for January 2022.
What is the probability that all 3 flights are on United? Give your answer as an exact decimal between 0 and 1 (not a Python expression).
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.
The average score on this problem was 93%.
What is the probability that all 3 flights are on Delta, or all on United, or all on American? Give your answer as an exact decimal between 0 and 1 (not a Python expression).
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.
The average score on this problem was 76%.
True or False: The probability that all 3 flights are on the same airline is equal to the probability you computed in the previous subpart.
True
False
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.
The average score on this problem was 90%.
King Triton has boarded a Southwest flight. For in-flight refreshments, Southwest serves four types of cookies – chocolate chip, gingerbread, oatmeal, and peanut butter.
The flight attendant comes to King Triton with a box containing 10 cookies:
The flight attendant tells King Triton to grab 2 cookies out of the box without looking.
Fill in the blanks below to implement a simulation that estimates the probability that both of King Triton’s selected cookies are the same.
# 'cho' stands for chocolate chip, 'gin' stands for gingerbread,
# 'oat' stands for oatmeal, and 'pea' stands for peanut butter.
cookie_box = np.array(['cho', 'cho', 'cho', 'cho', 'gin',
'gin', 'gin', 'oat', 'oat', 'pea'])
repetitions = 10000
prob_both_same = 0
for i in np.arange(repetitions):
grab = np.random.choice(__(a)__)
if __(b)__:
prob_both_same = prob_both_same + 1
prob_both_same = __(c)__What goes in blank (a)?
cookie_box, repetitions, replace=False
cookie_box, 2, replace=True
cookie_box, 2, replace=False
cookie_box, 2
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.
The average score on this problem was 92%.
What goes in blank (b)?
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.
The average score on this problem was 79%.
What goes in blank (c)?
prob_both_same / repetitions
prob_both_same / 2
np.mean(prob_both_same)
prob_both_same.mean()
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.
The average score on this problem was 93%.
You generate a three-digit number by randomly choosing each digit to be a number 0 through 9, inclusive. Each digit is equally likely to be chosen.
What is the probability you produce the number 027? Give your answer as a decimal number between 0 and 1 with no rounding.
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.
The average score on this problem was 92%.
What is the probability you produce a number with an odd digit in the middle position? For example, 250. Give your answer as a decimal number between 0 and 1 with no rounding.
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.
The average score on this problem was 78%.
What is the probability you produce a number with a 7 in it somewhere? Give your answer as a decimal number between 0 and 1 with no rounding.
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.
The average score on this problem was 69%.
The fine print of the Sun God festival website says “Ticket does not
guarantee entry. Venue subject to capacity restrictions.” RIMAC field,
where the 2022 festival will be held, has a capacity of 20,000 people.
Let’s say that UCSD distributes 21,000 tickets to Sun God 2022 because
prior data shows that 5% of tickets distributed are never actually
redeemed. Let’s suppose that each person with a ticket this year has a
5% chance of not attending (independently of all others). What is the
probability that at least one student who has a ticket cannot get in due
to the capacity restriction? Fill in the blanks in the code below so
that prob_angry_student evaluates to an approximation of
this probability.
num_angry = 0
for rep in np.arange(10000):
# randomly choose 21000 elements from [True, False] such that
# True has probability 0.95, False has probability 0.05
attending = np.random.choice([True, False], 21000, p=[0.95, 0.05])
if __(a)__:
__(b)__
prob_angry_student = __(c)__What goes in the first blank?
np.count_nonzero(attending) == 20001
attending[20000] == False
attending.sum() > 20000
np.count_nonzero(attending) > num_angry
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.
The average score on this problem was 67%.
What goes in the second blank?
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.
The average score on this problem was 59%.
What goes in the third blank?
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.
The average score on this problem was 68%.
You’re definitely going to Sun God 2022, but you don’t want to go alone! Fortunately, you have n friends who promise to go with you. Unfortunately, your friends are somewhat flaky, and each has a probability p of actually going (independent of all others). What is the probability that you wind up going alone? Give your answer in terms of p and n.
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.
The average score on this problem was 76%.
In past Sun God festivals, sometimes artists that were part of the lineup have failed to show up! Let’s say there are n artists scheduled for Sun God 2022, and each artist has a probability p of showing up (independent of all others). What is the probability that the number of artists that show up is less than n, meaning somebody no-shows? Give your answer in terms of p and n.
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.
The average score on this problem was 73%.
results = np.array([])
for i in np.arange(10):
result = np.random.choice(np.arange(1000), replace=False)
results = np.append(results, result)After this code executes, results contains:
a simple random sample of size 9, chosen from a set of size 999 with replacement
a simple random sample of size 9, chosen from a set of size 999 without replacement
a simple random sample of size 10, chosen from a set of size 1000 with replacement
a simple random sample of size 10, chosen from a set of size 1000 without replacement
Answer: a simple random sample of size 10, chosen from a set of size 1000 with replacement
Let’s see what the code is doing. The first line initializes an empty
array called results. The for loop runs 10 times. Each
time, it creates a value called result by some process
we’ll inspect shortly and appends this value to the end of the
results array. At the end of the code snippet,
results will be an array containing 10 elements.
Now, let’s look at the process by which each element
result is generated. Each result is a random
element chosen from np.arange(1000) which is the numbers
from 0 to 999, inclusive. That’s 1000 possible numbers. Each time
np.random.choice is called, just one value is chosen from
this set of 1000 possible numbers.
When we sample just one element from a set of values, sampling with replacement is the same as sampling without replacement, because sampling with or without replacement concerns whether subsequent draws can be the same as previous ones. When we’re just sampling one element, it really doesn’t matter whether our process involves putting that element back, as we’re not going to draw again!
Therefore, result is just one random number chosen from
the 1000 possible numbers. Each time the for loop executes,
result gets set to a random number chosen from the 1000
possible numbers. It is possible (though unlikely) that the random
result of the first execution of the loop matches the
result of the second execution of the loop. More generally,
there can be repeated values in the results array since
each entry of this array is independently drawn from the same set of
possibilities. Since repetitions are possible, this means the sample is
drawn with replacement.
Therefore, the results array contains a sample of size
10 chosen from a set of size 1000 with replacement. This is called a
“simple random sample” because each possible sample of 10 values is
equally likely, which comes from the fact that
np.random.choice chooses each possible value with equal
probability by default.
The average score on this problem was 11%.