Probability Examples 4

In this class, We discuss Probability Examples 4.

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The reader should have the basics of probability. Click Here.

Example 1:

In a certain town, 40% of the population has brown hair.

25% have brown eyes, and 15% have brown hair and brown eyes.

A person is selected randomly from a town.

1) if he has brown hair, what is the probability that he has brown eyes?

2) if he has brown eyes, is the probability that he does not have brown hair?

Solution:

Event A: Having brown hair

Event B: Having brown eyes.

P(A) = 40/100 = 0.4

P(B) = 25/100 = 0.25

P(A ∩ B) = 15/100 = 0.15

1) If he has brown hair probability of having brown eyes?

We go with conditional probability.

P(B|A) = P(A ∩ B) / P(A)

P(B|A) = (15/100) / (40/100)

P(B|A) = 3/8

2) if he has brown eyes, What is the probability of not having brown hair.

P(A’ |B) = P( A’ ∩ B)/ P(B)

A’ ∩ B = B – (A ∩ B).

The below Venn Diagram shows the violet region for A’ ∩ B.

Check our discrete mathematics course for sets and relations concepts.

P(A’ |B) = (P(B) – P(A ∩ B)) / P(B)

P(A’ |B) = ((25/100) – (15/100))/(25/100)

P(A’ |B) = 2/5

Example 2:

A class has ten boys and five girls.

Three students are selected at random, one after another.

Find the probability first and third are the same sex and the second one is the opposite sex.

Solution:

Possibilities: B1 G2 B3 or G1 B2 G3

The first one and third are boys and the second is a girl.

The first and third are girls and the second is a boy.

The two possibilities are mutually exclusive.

P(B1 G2 B3 U G1 B2 G3) = P(B1 G2 B3) + P(G1 B2 G3)

The question did not mention replacement. Default we are not replacing the student.

The selection is a dependent event because of not replacing the students.

P(B1∩ G2 ∩ B3) = P(B1) P(G2 |B1) P(B3 | (B1 ∩ G2))

P(B1 G2 B3 U G1 B2 G3) = P(B1∩ G2 ∩ B3) + P(G1∩ B2 ∩ G3)

P(B1 G2 B3 U G1 B2 G3) = (10/15 * 5/14 * 9/13) + (5/15 * 10/14 * 4/13)

P(B1 G2 B3 U G1 B2 G3) = 5/21