Question

Suppose that there is a white urn containing two white balls and three red balls and there is a red urn containing two white balls and three red balls. An experiment consists of selecting at random a ball from the white urn and then​ (without replacing the first​ ball) selecting at random a ball from the urn having the color of the first ball. Find the probability that the second ball is red.

Answer 1

The probability that the second ball is red is 11/25

Explanation:

Since the white urn contains 2 white balls and 3 red balls.

We have a total of 5 balls in this urn. The red urn contains 2 white balls and 3 red balls. We have a total of 5 balls likewise.

There are two ways of selecting red balls as the second.

1. Selecting white first, then selecting red

2. Selecting red first, then selecting red again.

IN THE EVENT OF SELECTING THE FIRST BALL

- The probability of selecting white ball is P(W) = 2/5

- Probability of selecting red ball is P(R) = 3/5

IN THE EVENT OF SELECTING THE SECOND BALL

P(R|W) = 1/2

P(R|R) = 3/5

So the probability of selecting a red ball as the second is

P(R2) = P(R|W)P(W) + P(R|R)P(R)

= (1/2)(2/5) + (2/5)(3/5)

= 1/5 + 6/25

= 11/25