Students in a college statistics class wanted to find out how common it is for young adults to have their ears pierced. They recorded data on two variables—gender and whether the student had a pierced ear—for all 178 people in the class. The two-way table below displays the data. Pierced Ears ?

Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

Suppose we choose a student from the class at random. Showing work and probability notation, find the probability that the student…1. has pierced ears.2. is a male with pierced ears.3. is male or has pierced ears.4. is female or has pierced ears.

1. 57.9%2. 10.7%

3. 97.8%

4. 60.1%

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

5.3 Conditional Probability and Independence (part 1)

“Probability of A given B”

Means the probability that event A occurs, given that event B has occurred◦ Or the probability that event A occurs, when we

know event B has occurred

Suppose we once again select a student at random from this college statistics class.

1. If we know that a randomly selected student has pierced ears, what is the probability that the student is male?

There are 103 students in the class with pierced ears, so we’ll restrict our attention to this group. Of those 103, how many are male?

|19103 18.4%

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

Suppose we once again select a student at random from this college statistics class.

2. If we know that a randomly selected student is male, what is the probability that the student has pierced ears?

This time, our attention is on the 90 male students in the class. Of those 90, how many have pierced ears?

1990 21.1%

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

So we have found that and

Notice that while these two questions sound alike, they are actually asking very different things (and the answers are almost always different).

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

Notice also that the unconditional probability of selecting a male student, , is different from the conditional probability of selecting a male if we know the student we selected has pierced ears.

18.4%, so knowing that extra piece of information changes our probability a great deal.

About half the students are male, but far less than half of the students with pierced ears are male.

Probability can change if we know additional information. That idea is what we call conditional probability.

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

The probability that event A and event B both happen is given by:

or alternately

Use the multiplication rule to find the percentage of students that are female and have pierced ears.

P(female and pierced) = P(female|pierced)*P(pierced)

Pierced Ears ?Gender Yes No TotalMale 19 71 90Female 84 4 88Total 103 75 178

For any two events A and B such that P(B)>0,

It’s on your formulas sheet as:

|∩

The multiplication rule is a consequence of this definition. Just solve for P(A and B)

Rolling two dice, use the definition of conditional probability to find the probability that you get a sum of 8 given that you rolled doubles.

Solution:◦ You already know the answer is , because there are six

equally likely ways to roll doubles. ◦ Using the definition:

Who Reads the Newspaper?Residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. 1. What is the probability that a randomly selected

resident who reads USA Today also reads the New York Times ?

2. What is the probability that a randomly selected resident who reads the New York Times does NOT read USA Today ?

Who Reads the Newspaper?Residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. 1. What is the probability that a randomly selected resident who reads USA Today also reads the New York Times?

There is a 12.5% chance that a randomly selected resident who reads USA Today also reads the New York Times.

∩ .. .

Who Reads the Newspaper?Residents of a large apartment complex can be classified based on the events A: reads USA Today and B: reads the New York Times. 2. What is the probability that a randomly selected resident who reads the New York Times does not read USA Today ?

There is an 80% chance that a randomly selected resident who reads the New York Times does NOT read USA Today.

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. .

Assignment #36◦ (due on Monday after Thanksgiving)

Read p. 318-323; do 57-60, 63, 66, 67, 71, 72