Probability Examples 1

In this class, we discuss Probability Examples 1.

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The reader should have prior knowledge of conditional probability. Click Here.

Example 1:

A bag contains ten gold and eight silver coins.

We made two successive draws of four coins.

1) The coins are replaced before the second draw.

2) The coins are not replaced before the second draw.

Find the probability that the first draw will give four gold coins and the second draw will give four silver coins.

1) With Replacement, the two events are independent.

Event A: First time, pick four gold coins

Event B: Second time, pick four silver coins.

P(A ∩ B) = P(A) P(B)

P(A) = 10c4/18c4

P(B) = 8c4/18c4

P(A ∩ B) = 10c4/18c4 * 8c4/18c4

2) Without Replacement, the two events are dependent.

For dependent events, we apply conditional probability.

Event A: First time, pick four gold coins

Event B: Second time, pick four silver coins.

P(A ∩ B) = P(A) P(B | A)

P(A) = 10c4/18c4

P(B | A) = 8c4/14c4

P(A ∩ B) = 10c4/18c4 * 8c4/14c4

Example 2:

Given a problem to three students A, B, and C.

The chances of solving the problem are 1/2, 3/4, and 1/4, respectively.

If any student solves the problem, We say it is a success.

What is the probability problem solved?

Solution:

Event A: Solve the problem.

Event B: Solve the problem.

Event C: Solve the problem.

The three events are independent of each other.

We need P(A U B U C)

P(A’) = 1- 1/2

P(B’) = 1 – 3/4

P(C’) = 1 – 1/4

P(A U B U C) = 1 – P((A U B U C)’)

P(A U B U C) = 1 – (P(A’ ∩ B’ ∩ C’))

P(A U B U C) = 1 – (P(A’) P(B’) P(C’)) = 1 – (1/2* 1/4 * 3/4)

P(A U B U C) = 29/32