A box has 10 red balls and 5 black balls A ball is selected

A box has 10 red balls and 5 black balls. A ball is selected from the box. If the ball is red, it is returned to the box. If the ball is black, it and 2 additional black balls are added to the box. Find the probability that a second ball selected from the box is

(a) red;

(b) black;

Solution

a)

Let

R1, R2 = the 1st, 2nd balls are red
B1, B2 = the 1st, 2nd balls are black

Hence, by Bayes\' Rule,

P(R2) = P(R1) P(R2|R1) + P(B1) P(R2|B1)

= (10/15)*(10/15) + (5/15)*(10/17)

= 0.640522876 [ANSWER]

*************

b)

By Bayes\' Rule,

P(B2) = P(R1) P(B2|R1) + P(B1) P(B2|B1)

= (10/15)*(5/15) + (5/15)*(7/17)

= 0.359477124 [ANSWER]