An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is

Let Ai(i=1,2,3,4,5) be the event that urn contains 2, 3, 4, 5 white balls and let B the event that two white balls have been drawn then we have to findP(A5/B) . Since the four events A2,A3,A4 and A5 are equally likely we have P(A2)=P(A3)=P(A4)=P(A5)=14 P(B/A2) is probability of event that the urn contains 2 white balls and both have been drawn.∴P(B/A2)=2C25C2=110 Similarly, ∴P(B/A3)=3C25C2=310 ,∴P(B/A4)=4C25C2=35 ,∴P(B/A5)=5C25C2=1 By baye’s theorem ,P(A5/B)=P(A5)P(B/A5)P(A2)P(B/A2)+P(A3)P(B/A3)+P(A4)(B/A4)+P(A5)P(B/A5)                    =14.114[110+310+35+1]=1020=12