One urn contains 3 red and 7 white marbles.

Question

A second urn contains 2 red and 8 white marbles. One urn is selected and a ball drawn. The probability of selecting the first urn is 0.4 and the probability of selecting the second is 0.6. If a white ball is drawn, find the probability it came from the second urn.

Answer 1

Probability of selecting first urn p(1) = 0.4 .

Probability of selecting second urn p(2) = 0.6 .

Total number of marbles in  first urn = 10 (3 red + 7 white)

Total number of marbles in  second urn = 10 (2 red + 8 white)

The probability of drawn a white ball from 2nd urn is

P(w/2)= (Number of white marbles in 2nd urn) / (Total number of marbles  in 2nd urn)

P(w/2)= 8 / 10 = 0.8 .

So the probability of Drawing a white ball from second urn is =p(2)*P(w/2) = 0.6*0.8 = 0.48 .

Hence the probability of white ball from second urn is 0.48 .