A bag contains n balls, one of which is white. The probability that A and B speak truth are `P_(1) and P_(2),` respectively. One ball is drwn from the bag and A and B both assert that it is white. Find the probability that drawn ball is actually white.

Explanation

[As n -1 balls remain in the bag and oen of them is white, the chancel that A should choose this ball and wrongly assert that it was drawn form the bas is `(1-P_(1))//(n-1).]`
Using total probability theorem,
`thereforeP(E)=P(A_(1))P(E//A_(1))+P(A_(2))P(E//A_(2))`
`=(P_(1)P_(2))/(n)+((n-1))/(n)((1-P_(1))(1-P_(2)))/((n-1)^(2))`
`=((n-1)P_(1)P_(2)+(1-P_(1))(1-P_(2)))/(n(n-1))`
Using Baye's teorem,
`impliesP(A_(1)//E)=(P(A_(1))P(E//A_(1)))/(P(E))`
`=((n -1)P_(1)P_(2))/((n-1)P_(1)P_(2)+(1-P_(1))(1-P_(2)))`