Computing Probabilities

STAT 20: Introduction to Probability and Statistics

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

Agenda

• Announcements
• PS 6
• Concept Review
• Concept Questions
• Break
• PS 7 (computing probabilities)
• Break

• PS 7 ~ 20/25 minutes: time for them to work, and to review the PS (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
• Handout: PS 8 (computing probabilities)

Announcements

• Problem Sets 6 and 7 (paper, max. 3) due Tuesday at 9am

• No lab this week.

• RQ: Probability Distributions due Mon/Tues 11:59pm

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

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

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. Show that 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 \(B^C\)

5. \(A^C\) and \(B\)

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.

03:00

02:00

The Houston Astros beat the Philadelphia Phillies in the 2022 World Series. 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 (best of 7). The Astros were heavily favored to win, so the outcome wasn’t really a suprise. Suppose we assumed that the probability that the Astros would have beaten the Phillies in any single game was estimated at 60%, independently of all the other games. What was the probability that the Astros would have won in a clean sweep?

(Clean sweep means that they won in the first 4 games - which didn’t happen, they won in 6 games.)

Straightforward application of multiplication rule for independent events. probability is 0.6^4 = 0.1296. 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 Astros would have beaten the Phillies 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

05:00

PS 7

25:00

Then review PS 7. If you have time, go over coin flipping code