Question

An urn contains 6 red balls and 3 blue balls. One ball is selected at random and is replaced by a ball of the other color. A second ball is then chosen. What is the conditional probability that the first ball selected is red, given that the second ball was red?

Answer 1

Answer:

0.5882 or 58.82%

Step-by-step explanation:

The probability that both balls were red (A) is:

$P(A)=\frac{6}{9}*\frac{5}{9}=0.3704$

The probability that the first ball was blue and the second ball was red (B) is:

$P(B) = \frac{3}{9}*\frac{7}{9}=0.2593$

The conditional probability that the first ball selected is red, given that the second ball was red is:

$P = \frac{P(A)}{P(A)+P(B)}=\frac{0.3704}{0.3704+0.2593} =0.5882$