Question

A bag A contains 4 black and 6 red balls and bag B contains 7 black and 3 red balls. A die is thrown. If 1 or 2 appears on it, then bag A is chosen, otherwise bag B. If two balls are drawn at random (without replacement) from the selected bag, find the probability of one of them being red and another black.

Answer 1

R for Red and B for Black.

Bag A ---- 4B and 6R,

Bag B ----7B and 3R,

If we get 1 or 2 then we select bag A. Thus probability of selection of bag A = P(A) =2/6

Probability of selection of bag B = P(B) = 4/6

Selection of balls could be RB, BR, RR, BB

For bag A, probability of one of balls being red and another black = RB + BR =4/10 x6/9+6/10  x 4/9  =48/90

For bag B, probability of one of balls being red and another black = RB + BR = 7/10 x3/9+3/10 x 7/9 =42/90

P = (selection of bag A) and (selection of RB or BR) or (selection of bag B ) and (selection of RB or BR)

P =2/6 x 48/90 +4/6 x 42/90

P = 7/15