Suppose that a box contains one 1 blue ball 2 red balls and

Suppose that a box contains one 1 blue ball, 2 red balls, and 3 white balls. suppose also that balls are selected from the box one at a time, at random, without replacement. What is the probability that the first ball is red given that the first red ball is obtained before the first white ball is obtained?

Solution

event A: first ball is red

event B: second ball is white

P(A|B) = P(AB)/P(B) = P(1st ball is red and 2nd ball is white)/P(2nd ball is white)

P(2nd ball is white) = P(1st ball is white)*P(2nd ball is white)+P(1st ball is not white)*P(2nd ball is white)

= (3/6)*(2/5) + ((1+2)/6)*3/5 = (6+9)/30 = 1/2

so,

P(A|B) = P(AB)/P(B) = P(1st ball is red and 2nd ball is white)/P(2nd ball is white)

= [(2/6)*(3/5)]/(1/2) = 6/(6*5*2) = 1/10 = 0.1