Conditional Probability

Chapter 15 - AP Statistics

some basic probability

terminology...

● Independent Events○ The outcome of one event

does not affect the other/next event

● Mutually Exclusive (Disjoint) Events○ Two events that do not

have any outcomes in common (or, two events that cannot both happen)

**If two events are DISJOINT, they cannot be INDEPENDENT!

some basic probability

notation

Let’s say we have two “events” that we’ll call “A” and “B” ...

● Probability of “A” and “B”

P(A ⋂ B)

● Probability of “A” or “B”

P(A ⋃ B)

● Probability of “A” given “B”

P(A | B)

some basic probability formulas…

Conditional Probability

(these are on your formula chart)

P(A | B) = P(A ⋂ B)

P(B)

Probability of “A” or “B”

P(A ⋃ B) = P(A) + P(B) - P(A ⋂ B)

let’s talk about probability of “A” and “B”...

If “A” and “B” are independent…

P(A ⋂ B) = P(A) x P(B)

Otherwise…

P(A ⋂ B) = P(A) x P(B | A)

OR

P(A ⋂ B) = P(B) x P(A | B)

IMPORTANT to REMEMBER!

P(A ⋂ B) ≠ P(A) x P(B)

does not equal

unlessevents “A” and “B” are independent!!

If they are NOT independent, perhaps P(A ⋂ B) was given in the problem, or there’s a way to figure out the conditional probability.

1. Probability & Wastepaper Basketball

Two-way table: Gender vs. Making the Shot?

Made it

Missed it

Guys 5 8

Girls 9 7

Total

13

16

29Total 14 15

a) What is the probability that a shot made it in the trash can?

P(shot made) = = 0.4828

b) What is the probability that a shot was taken by a female?

P(female) = = 0.5517

c) What is the probability that a shot made it in the trash can, and was taken by a female?

P(shot made ⋂ female) = = 0.3103

1429

1629

Look at the table!!!!

Don’t multiply!!9

29

d) What is the probability that a shot made it in the trash can OR was taken by a female?

P(shot made ⋃ female) = + - = = 0.7241

e) Given that a shot was made, what is the probability the person who threw it was male?

P(male | shot made) = = 0.3571

f) What is the probability that a female shooter made the shot?

P(shot made | female) = = 0.5625

1429

1629

929

2129

Look at the table!!!! 5

14

Look at the table!!!!

916

g) Are “making the shot” and “being male” independent? Justify your answer. Test for Independence: does P(A) = P(A | B)?

P(shot made) ? P(shot made | male)

0.4827 ≠ 0.3846

h) Assuming the probabilities remain true for future shots taken, what is the probability that 2 shots in a row make it in the trash can?

P(2 shots made) = x = 0.2331

1429

513

No, they are not independent, since P(shot made) ≠ P(shot made | male). (Males are LESS likely to make the shot than the class as a whole.)

1429

1429

2. The Dorm Room Problem

2. The Dorm Room ProblemA check of dorm rooms on a large college campus revealed that 38% of the rooms had refrigerators, 53% had TVs, and 21% had both a refrigerator and a TV. helpful to write out what you HAVE to start with: P(refrigerator) = 0.38 P(TV) = 0.53 P(refrigerator ∩ TV) = 0.21

a) What is the probability that a randomly selected dorm room has a TV or a refrigerator?

P(TV ∪ fridge) = P(TV) + P(fridge) - P(both) = 0.38 + 0.53 - 0.21 = 0.70

b) What is the probability that a dorm room with a refrigerator also has a TV?

P(TV | fridge) = P(TV ∩ fridge) = 0.21 = 0.5526P(fridge) 0.38

2. The Dorm Room ProblemA check of dorm rooms on a large college campus revealed that 38% of the rooms had refrigerators, 53% had TVs, and 21% had both a refrigerator and a TV.

P(refrigerator) = 0.38 P(TV) = 0.53 P(refrigerator ∩ TV) = 0.21

c) What is the probability that a dorm room with a TV also has a refrigerator?

P(fridge | TV) = P(fridge ∩ TV) = 0.21 = 0.3962 P(TV) 0.53

d) Are the events “has a TV” and “has a refrigerator” mutually exclusive (or, disjoint)? Explain.

No, because 21% of dorm rooms had both a refrigerator and a TV. If these were disjoint, it would be impossible for both to happen.

2. The Dorm Room ProblemYou can also make a Venn diagram to help visualize where all the probabilities are!

refrigerator tv

21%17% 32%

30%

Fridge ONLY TV ONLYboth

neither

3. The Professor Problem

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

Professor Scedastic

Professor Kurtosis

Passed

Failed

Passed

Failed

0.58

0.42

0.8

0.2

0.6

0.4

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

Professor Scedastic

Professor Kurtosis

Passed

Failed

Passed

Failed

0.58

0.42

0.8

0.2

0.6

0.4

P(Sced ⋂ Pass) =

P(Sced ⋂ Fail) = 0.58 x 0.2 = 0.116

P(Kurt ⋂ Pass) = 0.42 x 0.6 = 0.252

P(Kurt ⋂ Fail) = 0.42 x 0.4 = 0.168

0.58 x 0.8 = 0.464P(Scedastic)

P(Passed | Scedastic))

x

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

a) What is the probability that Molly gets Professor Scedastic, AND passes the class?

P(Scedastic ⋂ Pass) = 0.464

b) What is the probability that Molly passes the class?

P(Pass) = add up both scenarios when Molly could pass = P(Scedastic ∩ Pass) + P(Kurtosis ∩ Pass) = 0.464 + 0.252 = 0.716

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

c) In Professor Kurtosis’s class, what proportion of the students didn’t pass the course?

P(Failed | Kurtosis) = 0.4 (you can look at your tree diagram, OR use the conditional formula)

d) In Professor Scedastic’s class, what proportion of the students DID pass the course?

P(Passed | Scedastic) = 0.8 (you can look at your tree diagram, OR use the conditional formula)

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

c) In Professor Kurtosis’s class, what proportion of the students didn’t pass the course?

P(Failed | Kurtosis) = 0.4

d) In Professor Scedastic’s class, what proportion of the students DID pass the course?

P(Passed | Scedastic) = 0.8

e) What is the probability that a student who PASSED the course had Professor Scedastic?

P(Scedastic | Passed) = … hmm

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

Professor Scedastic

Professor Kurtosis

Passed

Failed

Passed

Failed

0.58

0.42

0.8

0.2

0.6

0.4

P(Sced ⋂ Pass) =

P(Sced ⋂ Fail) = 0.116

P(Kurt ⋂ Pass) = 0.252

P(Kurt ⋂ Fail) = 0.168

0.464

There’s no branch in our tree

diagram that holds ALL the

people who passed the class!

3. The Professor Problem (tree diagrams)Molly’s college offers two sections of Statistics 101. Since students are randomly assigned to the two classes, she knows that she has a 58% chance of getting Professor Scedastic, who has taught some of her friends in the past. Based on what she’s heard about the two professors, Molly estimates that her chances of passing the course are 80% if she gets Professor Scedastic, but only 60% if she gets Professor Kurtosis.

c) In Professor Kurtosis’s class, what proportion of the students didn’t pass the course?

P(Failed | Kurtosis) = 0.4

d) In Professor Scedastic’s class, what proportion of the students DID pass the course?

P(Passed | Scedastic) = 0.8

e) What is the probability that a student who PASSED the course had Professor Scedastic?

P(Scedastic | Passed) = P(Scedastic ⋂ Passed)

P(Passed)= 0.464

0.464 + 0.252= 0.648