# MATH 102 Spring 2011 Extra Examples

Here are a few more sample problems with solutions, to supplement the examples in the textbook, the homework problems, and the examples shown in lecture.

## Conditional Probability and Independence

1. Say that in a certain town 1/3 of the people vote Conservative, and the probability that a townsperson drives a car more than ten years old is 3/11. If you pick a random townsperson, the probability that that person is a Conservative driving a car more than ten years old is 12%. In this town, is voting Conservative independent of driving a car more than ten years old? Why or why not?

No, not independent.
Let A= Conservative, B = old car.
P(A) = 1/3; P(B) = 3/(3+8) = 3/11.
P(A)P(B) = 1/11, but P(A and B) = 12%: not the same, so A and B are not independent.

In words, if voting Conservative were independent of driving an old car, then we could calculate the chance that a random person is both Conservative and driving an old car (the intersection) by multiplying the probability of being Conservative (1/3) by the probability of driving an old car (3/11). But this doesn't yield the right number, so the two are not independent.

A common confusion is to say, "Well, just because a townsperson is Conservative doesn't mean they can't drive an old car: hence the two are not independent.". That is not a valid conclusion; that statement has more to do with overlap (mutual exclusivity) than with independence. But independence deals with probabilities: does the probability of voting Conservative depend on whether they drive an old car or not?

2. Let A represent students with financial aid. Let B represent students from BC. Translate each of the following statements into P() notation and draw Venn diagrams in the boxes provided, shading in each region described:
1. 39% of students are neither from BC nor have financial aid.

P( (A or B)C ) = 39% (complement of union)

2. 12% of all students are from BC and also have financial aid.

P(A and B) = 12% (intersection)

3. 25% of BC students have financial aid. (No Venn diagram needed.)

P(A | B) = 25% (if you had to draw a Venn diagram, it'd be a fraction with two Venn diagrams, one for the numerator and one for the denominator.)

4. Given this information, what percentage of students are from BC?

P(B) = P(A and B) / P(A | B) = 12% / 25% = 4 * (12%) = 48%

5. What fraction of students have financial aid?

P(A) = (1- P((A or B)C)) - P(B) + P(A and B)
= (1 - 39%) - 48% + 12%
= 25%

6. Are the two events A and B independent? Why or why not?

Yes, independent:
P(A) = 25% = P(A | B)

7. Are the two events A and B mutually exclusive? Why or why not?

No: P(A and B) = 12% is not zero.

3. Two researchers, Dr. Prudence and Dr. Folly, are often asked by pharmaceutical companies to evaluate new experimental cancer drugs. The pharmaceutical companies often invite the researchers to lavish conferences in exotic destinations.
1. For companies which are extravagant, Dr. Folly approves the drugs 85% of the time. If the company is not extravagant, Dr. Folly approves the drugs only 70% of the time.

Define the sample space and two events A and B, and translate the preceding two sentences into probability notation using A and B.

Sample space: drugs evaluated by Dr. Folly.
Event A: drugs approved by Dr. Folly
Event B: drugs produced by an extravagant company
P (A | B) = 0.85 (A conditioned on B)
P (A | BC) = 0.70 (A conditioned on "not B")

2. Are the two events A and B independent? Why or why not?

Not independent: P(A|B) is not equal to P(A|BC), hence P(A|B) cannot be equal to P(A). P(A) is the probability that Dr. Folly will approve a drug, regardless of company.

3. Dr. Prudence approves only 20% of the drugs evaluated. Even if the pharmaceutical company is extravagant, Dr. Prudence still only approves the drugs 20% of the time. Define a sample space and events A and B, and translate the preceding two sentences into probability notation using A and B.

Sample space: drugs evaluated by Dr. Prudence.
Event A: drugs approved by Dr. Prudence
Event B: drugs produced by an extravagant company
P (A) = 0.20
P (A | B) = 0.20 (A conditioned on B)

4. Are the two events A and B independent? Why or why not?

Independent: P(A)=P(A|B).

5. Interpret these results in the context of the moral ethics of the two researchers.

Dr. Folly's approval of a drug depends on whether the drug company is extravagant. The chance that Dr. Prudence approves a drug is independent of whether the drug company is extravagant.