Probability Foundations II

STAT 20: Introduction to Probability and Statistics

Agenda

• Announcements
• Review: Probability Foundations I
• Break
• Lecture: Probability Foundations II
• Appendix: More practice!

Announcements

• Portfolio 3 due next Monday. The Gradescope assignment will go up on Thursday.

• Quiz 2 is next Thursday and covers everything from Monday to next Wednesday.

Review: Probability Foundations I

30:00

Break

05:00

Lecture Probability Foundations II

Appendix: More practice!

Do you want a car or do you want a goat?

Of course, this is the Monty Hall problem. The text for this comic is “A few minutes later, the goat behind door C drives away in the car. Can have some fun and ask them to if they chose door A and door B had a goat, would they stay with A or would they switch to C? (pollev) Setup: There are three doors A, B, C. Car behind one of them, goats behind other two. You pick a door at random (say you pick A), host opens one of the doors you didn’t pick (say they open B), and shows you a goat. Now should you stick with your original choice of A or should you switch to C. Will be fun to see what they input. The right answer is that they should switch. Waving hands, the total mass of 1 was split 1/3 to 2/3 (win to not win) with the choice of A. A is kept aside, and B is opened.Now the P(A doesn’t win) = 2/3 is moved to C.Therefore better to switch. Not worth doing it with formal notation even after going over Bayes. From Everett’s suggestions (spring 2025): Have the class”play” this game. Explain the scenario, draw “doors” on the board, maybe even invite a “contestant,” and have a poll to ask “should they switch?” with options (yes | no | it doesn’t matter). Then show (through code/simulation, then math) why switching is better. For conditional probability, one good example to share is election polling. Trump/Harris was pretty much 50/50 by all reasonable pollsters, but then trump won by kind of a lot. Among other reasons, this is because the “swing states” probabilities were not independent - there was some latent unknown “trumpiness” that affected all of them.

Birthday problem

This is a fun one we will play out in class! What is the probability that with \(K\) people in a room, at least 2 of them share a birthday? It turns out that with just 23 people there is already about a 50% chance two people share a birthday.

Check your answer

It is far easier to compute the complement — the probability that everyone’s birthday is different — and subtract from 1.

Line the people up. The 1st person can have any birthday. The 2nd must avoid that one day: \(\tfrac{364}{365}\). The 3rd must avoid two days: \(\tfrac{363}{365}\), and so on. Multiplying: \[P(\text{all different}) = \frac{365}{365}\cdot\frac{364}{365}\cdots\frac{365-K+1}{365} = \frac{365!}{(365-K)!\,\;365^{K}}.\]

So the probability of at least one shared birthday is \[P(\text{at least one match}) = 1 - \frac{365!}{(365-K)!\,\;365^{K}}.\]

Plugging in values gives the (famously surprising) numbers:

People \(K\)	10	23	30	50	57	70
\(P(\text{match})\)	0.117	0.507	0.706	0.970	0.990	0.999

At \(K = 23\) we cross 50%, and by \(K = 70\) a shared birthday is all but certain. The surprise comes from the fact that we are comparing all pairs of people, and the number of pairs grows fast — there are \(\binom{23}{2} = 253\) pairs among just 23 people.

(Assumptions: 365 equally likely birthdays, no leap years, no twins.)

Concept questions & review

Rules

• Conditional Probabilty

For two events \(A\) and \(B\), \(P(A \vert B) = \displaystyle \frac{P(A \text{ and } B)}{P(B)}\)

• Multiplication rule

For two events \(A\) and \(B\), \(P(A \text{ and } B) = P(A \vert B) P(B)\)

• Complement rule

\(P(A^C) = 1 - P(A)\)

• WS needs 20/25 minutes: time for them to work, and for you to review the problems (If you have already given enough time, do the coin flipping activity)
• Lecture on conditional probability and independence
• Concept Review
• Concept questions
• Break ~3 mins

You can tell them here that \(A^C\) is “not A”, and then maybe use dice or coin toss examples. Review mutually exclusive vs independent events, emphasizing that these are very different ideas, though both apply to pairs of events.

• Mutually exclusive events means that the occurrence of one event prevents the occurrence of the other. (that is, it reduces the chance of the other occurring to 0.)

Independent events means that the occurrence of one event does not change the chance of the other occurring.

Concept Question 1

01:00

Flip 3 coins, one at a time. Define the following events:

\(A\) is the event that the first coin flipped shows a head

\(B\) is the event that the first two coins flipped both show heads

\(C\) is the event that the last two coins flipped both show tails

The events \(A\) and \(B\) are: ________

After they do this and next, write out the outcome space for the results of flipping 3 coins. And go through what are A, B, and C.

A and B are neither independent nor mutually exclusive.

Concept Question 2

01:00

Flip 3 coins, one at a time. Define the following events:

\(A\) is the event that the first coin flipped shows a head

\(B\) is the event that the first two coins flipped both show heads

\(C\) is the event that the last two coins flipped both show tails

The events \(A\) and \(C\) are: ________

After they do this and next, write out the outcome space for the results of flipping 3 coins. And go through what are A, B, and C.

A and C are independent. P(A & C ) = 1/8 = P(A)P(C) = 1/4 * 1/2

Concept Question 3

Suppose we draw 2 tickets at random without replacement from a box with tickets marked {1, 2, 3, . . . , 9}. Let \(A\) be the event that at least one of the tickets drawn is labeled with an even number, let \(B\) be the event that at least one of the tickets drawn is labeled with a prime number (recall that the number 1 is not regarded as a prime number). Suppose the numbers on the tickets drawn are 3 and 9.

Which of the following events occur?

1. \(A\)

2. \(B\)

3. \(A\) and \(B\) (\(= A \cap B\))

4. \(A\) and \(B^c\) (\(= A \cap B^c\))

5. \(A^c\) and \(B\) (\(= A^c \cap B\))

ii and v, I told them they can immediately see that A^C is true since no even number, so definitely (v) and then we do have a prime number and both are odd so (ii) is true.

You could give other events for practice such as 2 and 9, or 4 and 9 etc.

02:00

01:00

The 2024 World Series was the championship series was a best-of-seven playoff between the Los Angeles Dodgers and the New York Yankees. The winners in the World Series have to win a majority of 7 games, so the first team to win 4 games wins the series. Suppose we assumed that the probability that the Dodgers would have beaten the Yankees in any single game was estimated at 56%, independently of all the other games.

What was the probability that the Dodgers would have won in a clean sweep?

It was the Dodgers’ first World Series appearance and win since 2020, and the Yankees’ first World Series appearance since 2009. Straightforward application of multiplication rule for independent events. probability is 0.56^4 = 0.098. You could even ask more or less than 10%?

You may have to explain Clean sweep means that they won in the first 4 games - which didn’t happen, they won in 5 games.

Follow up comment/question: if the teams were more evenly matched, say probability of either team winning is 50%, the probability of going to 7 games and probability of winning in 6 games is equal given 5 games are played.

Concept Question 5

01:00

Suppose we assume, instead, that the probability that the Dodgers would have beaten the Yankees in any single game was 50%, independently of all the other games. In this case, was the probability that the series would have gone to 6 games higher than the probability that the series would have gone to 7 games, given that 5 games were played?

if 5 games are played it means one of the teams is leading 3-2. Therefore there are two scenarios: the team that is ahead wins game 6 and the series, or the team that is behind wins game 6 and they go to 7 games. Both are equally likely.

Break

03:00

Worksheet: Computing Probabilities

25:00

Then review the WS