CSE 312 – Section 3

Spring 2026

Review of Main Concepts

Conditional Probability: $\Pr (A|B) =\dfrac {\Pr (A \cap B)}{\Pr (B)}$
Independence: Events $E$ and $F$ are independent iff $\Pr (E \cap F)$ = $\Pr (E)\Pr (F)$, or equivalently $\Pr (F) =\Pr (F|E)$, or equivalently $\Pr (E) =\Pr (E|F)$
Bayes Theorem: $\Pr (A|B) =\dfrac {\Pr (B|A)\Pr (A)}{\Pr (B)}$
Partition: Nonempty events $E_1,\dots ,E_n$ partition the sample space $\Omega $ iff
- $E_1,\dots ,E_n$ are exhaustive: $E_1 \cup E_2 \cup \dots \cup E_n=\bigcup _{i=1}^n E_i =\Omega $, and
- $E_1,\dots ,E_n$ are pairwise mutually exclusive: $\forall i \neq j, E_i \cap E_j=\emptyset $
Law of Total Probability (LTP): Suppose $A_{1},\ldots ,A_{n}$ partition $\Omega $ and let $B$ be any event. Then

$\Pr ( B ) = \sum _{i = 1}^{n}{\Pr (B \cap A_{i})} = \sum _{i = 1}^{n}{\Pr ( \text {B\ } |\ A_{i})\Pr (A_{i})}$
Bayes Theorem with LTP: Suppose $A_{1},\ldots ,A_{n}$ partition $\Omega $ and let $B$ be any event. Then $ \Pr ( A_1 | B) =\dfrac {\Pr ( B\ |\ A_1 ) \Pr (A_1)}{\sum _{i = 1}^{n}{\Pr ( \text {B\ } |\ A_{i})\Pr (A_{i})}} $. In particular, $ \Pr ( A | B) = \dfrac {\Pr ( B\ |\ A )\Pr ( A ) }{\Pr ( \text {B\ } |\ A)\Pr ( A ) + \Pr ( \text {B\ } |\ A^{C})\Pr (A^{C})} $
Chain Rule: Suppose $A_1,...,A_n$ are events. Then, \[\Pr (A_1\cap ...\cap A_n)=\Pr (A_1)\Pr (A_2|A_1)\Pr (A_3|A_1\cap A_2)...\Pr (A_n|A_1\cap ...\cap A_{n-1})\]
Conditional Independence: Events $E$ and $F$ are independent conditioned on event $G$ iff $\Pr (E \cap F|G)$ = $\Pr (E|G)\Pr (F|G)$.

Section Plan

Administrivia

Pset 1 grades released on Gradescope. Check your submissions to read comments. Regrade requests open 24 hours after grades are released and close after a week
PSet 2 was due yesterday 4/15 @ 11:59 PM. Late deadline Saturday @ 11:59 PM (max of 3 late days per problem set)
Pset 3 is out and due next Wednesday (April 22). Submit on Gradescope.
Pset 3 has a coding part! you will be working partly in Ed, but see instructions on PSet
For an introduction to the Naive Bayes Classifier and details on implementation, see the video and slides (as well as some self-check questions) linked at this edstem lesson. Also check out Section 9.3 from the book. To solve the task, we have set up an edstem lesson.

Questions about material covered so far?

Problems to be covered in section

problem 1
problem 3
problem 5
problem 9
Time permitting, problem 10 and 11.

Be sure to check out all the remaining problems before you do your homework.

1 Content Review

a)

True or False: It is always the case that $\Pr (A \mid B) = \Pr (B \mid A)$.

b)

Select one: Suppose $A$ and $B$ are independent events. Then

$\Pr (A \cap B) = \Pr (A)\Pr (B)$
$\Pr (A \mid B) = \Pr (A)$
Both are true
Both are false

c)

Select one: For any two events $A$ and $B$,

$\Pr (A \mid B) = \Pr (A \cap B)$
$\Pr (A \mid B) = \frac {\Pr (B \mid A)\Pr (A)}{\Pr (B)}$
$\Pr (A \mid B) = \frac {\Pr (B \mid A)\Pr (B)}{\Pr (A)}$

d)

True or False: Let $A$ and $B$ be the event that a roll of a six-sided die shows a number that is at most 3 or more than 4, respectively. Then $A$ and $B$ are a partition.

e)

Select one: Let $T$ be the event that Alice tests positive for the flu. Let $C$ be the event that Alice actually has the flu. Suppose the probability that Alice tests positive given that she has a flu is $0.8$. Then the probability she tests negative given that she has a flu is

$0.8$
$0.2$
Not enough information

f)

True or False: Suppose $A_1,\ldots ,A_n$ are events. Then \[ \Pr (A_1 \cap A_2 \cap \cdots \cap A_n) = \Pr (A_1) \cdot \Pr (A_2 \mid A_1) \Pr (A_3\mid A_1 \cap A_2) \cdots \Pr (A_n \mid A_1 \cap \cdots \cap A_{n-1})\;. \]

2 Flipping Coins

A coin is tossed twice. The coin is “heads” one quarter of the time. You can assume that the second toss is independent of the first toss.

a): What is the probability that the second toss is “heads” given that the first toss is “tails”?
b): What is the probability that the second toss is “heads” given that at least one of the tosses is “tails”?
c): With respect to the probability space of this problem, give an example of two events that are disjoint but not independent.
d): With respect to the probability space of this problem, give an example of two events that are independent but not disjoint.

3 Balls from an Urn

Say an urn contains three red balls and four blue balls. Imagine we draw three balls without replacement. (You can assume every ball is uniformly selected among those remaining in the urn.)

a): What is the probability that all three balls are all of the same color?
b): What is the probability that we get more than one red ball given the first ball is red?

4 Random Grades?

Suppose there are three possible teachers for CSE 312: Martin Tompa, Anna Karlin, and Adam Blank. Suppose the probability of getting an $A$ in Tompa’s class is $\frac {5}{15}$, in Karlin’s class is $\frac {3}{15}$, and in Blank’s class is $\frac {1}{15}$. Suppose you are assigned a grade randomly according to the given probabilities when you take a class from one of these professors, irrespective of your performance. Furthermore, suppose Blank teaches your class with probability $\frac {1}{2}$ and Karlin and Tompa have an equal chance of teaching if Blank isn’t. Use Bayes’ Rule to compute the probability you had Blank, given that you received an $A$. Compare this to the unconditional probability that you had Blank.

5 Marbles in Pockets

Aleks has 5 blue and 3 white marbles in her left pocket, and 4 blue and 4 white marbles in her right pocket. If she transfers a randomly chosen marble from left pocket to right pocket without looking, and then draws a randomly chosen marble from her right pocket, what is the probability that it is blue?

6 Game Show

Corrupted by their power, the judges running the popular game show America’s Next Top Mathematician have been taking bribes from many of the contestants. During each of two episodes, a given contestant is either allowed to stay on the show or is kicked off. If the contestant has been bribing the judges, she will be allowed to stay with probability 1. If the contestant has not been bribing the judges, she will be allowed to stay with probability 1/3, independent of what happens in earlier episodes. Suppose that 1/4 of the contestants have been bribing the judges. The same contestants bribe the judges in both rounds.

a): If you pick a random contestant, what is the probability that she is allowed to stay during the first episode?
b): If you pick a random contestant, what is the probability that she is allowed to stay during both episodes?
c): If you pick a random contestant who was allowed to stay during the first episode, what is the probability that she gets kicked off during the second episode?
d): If you pick a random contestant who was allowed to stay during the first episode, what is the probability that she was bribing the judges?

7 Parallel Systems

A parallel system functions whenever at least one of its components works. Consider a parallel system of $n$ components and suppose that each component works with probability $p$ independently.

a): What is the probability the system is functioning?
b): If the system is functioning, what is the probability that component 1 is working?
c): If the system is functioning and component 2 is working, what is the probability that component 1 is also working?

8 Allergy Season

In a certain population, everyone is equally susceptible to colds. The number of colds suffered by each person during each winter season ranges from 0 to 4, with probability 0.2 for each value (see table below). A new cold prevention drug is introduced that, for people for whom the drug is effective, changes the probabilities as shown in the table. Unfortunately, the effects of the drug last only the duration of one winter season, and the drug is only effective in 20% of people, independently.


number of colds	no drug or ineffective	drug effective

0	0.2	0.4
1	0.2	0.3
2	0.2	0.2
3	0.2	0.1
4	0.2	0.0

a): Sneezy decides to take the drug. Given that he gets 1 cold that winter, what is the probability that the drug is effective for Sneezy?
b): The next year he takes the drug again. Given that he gets 2 colds in this next year (and one cold in the first year), what is the probability that the drug is effective for Sneezy?
c): Why is the answer to (b) the same as the answer to (a)?

9 What’s the chance it was raining?

Suppose that you can commute to campus via bike or public transit. If it’s raining, you’ll bike to campus $20\%$ of the time, and if it’s not raining, you’ll bike to campus $60\%$ of the time. You know that it rains 7 days out of 10 during Autumn quarter. If you biked to campus today, what is the probability that it was raining?

10 A game

Howard and Jerome are playing the following game: A 6-sided die is thrown and each time it’s thrown, regardless of the history, it is equally likely to show any of the six numbers.

If it shows 5, Howard wins.
If it shows 1, 2, or 6, Jerome wins.
Otherwise, they play a second round and so on.

What is the probability that Jerome wins on the 4th round?

11 Another game

Alice and Alicia are playing a tournament in which they stop as soon as one of them wins $n$ games. Alicia wins each game with probability $p$ and Alice wins with probability $1-p$, independently of other games. What is the probability that Alicia wins and that when the match is over, Alice has won $k$ games?

12 Dependent Dice Duo

This problem demonstrates that independence can be “broken” by conditioning. Let $D_1$ and $D_2$ be the outcomes of two independent rolls of a fair die. Let $E$ be the event “$D_1 = 1$”, $F$ be the event “$D_2 = 6$”, and $G$ be the event “$D_1 + D_2 = 7$”. Even though $E$ and $F$ are independent, show that \[ \mathbb {P}(E \cap F \mid G) \neq \mathbb {P}(E \mid G) \cdot \mathbb {P}(F \mid G) \]

13 Infinite Lottery

Suppose we randomly generate a number from the natural numbers $\mathbb {N} = \{ 1,2,\ldots \}$. Let $A_{k}$ be the event we generate the number $k$, and suppose $\Pr ( A_{k} ) = \left ( \frac {1}{2} \right ) ^{k}$. (Note that $\sum _{k=1}^\infty 2^{-k} = 1$.)

Once we generate a number $k$, that is the maximum we can win. That is, after generating a value $k$, we can win any number in $[k]=\{1,...,k\}$ dollars. Suppose the probability that we win $\$ j$ for $j\in [k] $ is “uniform”, that is, each has probability $\frac {1}{k}$. Let $B$ be the event we win exactly $\$ 1$. Given that we win exactly one dollar, what is the probability that the number generated was also $1$?

You may use the fact that $\sum _{j = 1}^{\infty }\frac {1}{j \cdot a^{j}} = \ln \left ( \frac {a}{a - 1} \right ) $ for $a > 1$.

14 The mysteries of independence (10 points)

Suppose that a uniformly random card is selected from a standard 52 card deck of cards. Let $E$ be the event that the card is a king, let $F$ be the event that the card is a heart, and let $G$ be the event that the card is black (that is, a spade or a club).

a)

(i): Are $E$ and $F$ independent? Provide a short proof of your claim.
(ii): Are $G$ and $F$ independent? Provide a short proof of your claim.
(iii): Are $E$ and $G$ independent? Provide a short proof of your claim.

b)

Now assume that an additional green card (with no suit and no rank) is added to the deck and a uniformly random card is selected from this enlarged deck of 53 cards. Let $E'$ be the event that the card is a king, let $F'$ be the event that the card is a heart, and let $G' $ be the event that the card is black (that is a spade or a club).

(i): Are $E'$ and $F'$ independent? Provide a short proof of your claim.
(ii): Are $G'$ and $F'$ independent? Provide a short proof of your claim.
(iii): Are $E'$ and $G'$ independent? Provide a short proof of your claim.

A proof that two events $A$ and $B$ are independent typically consists of showing that $Pr(A \cap B) = Pr (A ) \cdot Pr (B)$, whereas a proof that they are not independent consists of showing that $Pr(A \cap B) \ne Pr (A ) \cdot Pr (B)$

15 CSE 312 Learning Assistant Usage

There is a population of $n$ students in the class. The number of students using the 312 Learning Assistant among these $n$ students is $i$ with probability $p_i$, where, of course, $\sum _{0\le i \le n}p_i = 1$. We take a random sample of $k$ students from the class (without replacement). Use Bayes Theorem to derive an expression for the probability that there are $i$ students using the 312 Learning Assistant in the population conditioned on the fact that there are $j$ such students in the sample. Let $G_i$ be the event that there are $i$ students using the 312 Learning Assistant in the population and let $S_j$ be the event that there are $j$ such students in the sample. Your answer should be written in terms of the $p_\ell $’s, $i,j,n$ and $k$.

16 The Monty Hall Problem

The Monty Hall problem is a famous, seemingly counter-intuitive probability puzzle named after Monty Hall, the host of the show "Let’s Make a Deal". This problem emphasizes the importance of using given information to make decisions. Assume you are a contestant on this game show. In the original problem, there are $3$ doors, one (randomly selected) hiding a car and the other two hiding goats. At first, you randomly pick a door, hoping you can win the car. As Monty knows exactly what door hides the location of the car, he purposefully decides to reveal a door different from your pick which is guaranteed to reveal a goat. As there are $2$ doors left, Monty later asks if you want to stick to your current door or to switch to the other door.

In the beginning, when there is no information about these $3$ doors, every door has equal probability of revealing a car. However, after knowing that Monty will only open a door which definitely reveals a goat, it turns out that switching to the other door yields a higher probability of winning than sticking to your current door. Thus, the best strategy is to switch to the other door.

For this problem, you have to determine the best strategy when there are $4$ doors, again a random one hiding a car and the other 3 hiding one goat each. As a contestant, you first randomly choose a door. Monty opens one of the $3$ other doors, which reveals a goat, and asks if you want to stick to your current choice or switch to a different door. After you make your pick, Monty opens another door (other than your current pick) which also reveals a goat. This time, you have to make the final pick: sticking to the current door in the previous pick or switching to the other door. Perform a thorough analysis of all possible strategies and explain which one is the best.

17 More independence

Prove that if events $E$ and $F$ are independent, then so are $E$ and $\overline {F}$.