From an urn containing `a` white and `b` black balls, `k` balls are drawn and laid aside, their color unnoted. Then one more ball is drawn. Find the probability that it is white assuming that k

Explanation

Let `E_(1)` denote the event that out of the first k balls drawn, I balls are ehite, and A be the event that `(k+1)^(1)` ball drawn is also white. We have to find P(A). Now ways of selecting I white balls feom a white balls and k-I black balls from b black balls is `""^(a)C_(i)""^(b)C_(k-i)(0leilek).` ways to sleck k balls from a + b balls is `""^(a+b)C_(k)` Therefore,
`P(E_(i))=(""^(a)C_(i)""^(b)C_(k-i)),0leilek`
Also, `P(A//E_(i))=(""^(a-i)C_(1))/(""^(a+b-k)C_(1))=(a-i)/(a+b-k)(0leilek)`
By the theorem of total probability, we have
`P(A)=underset(i-0)overset(k)sumP(E_(i))P(A//E_(i))` ltlbrgt `=underset(i-0)overset(k)sum (""^(a)C_(i)""^(b)C_(k-i))/(""^(a+b)C_(k))(a-1)/(a+b-k)`
`=underset(i-0)overset(k)sum([(a-i)""^(a)C_(a-i)]""^(b)C_(k-i))/((a+b-k)^(a+b)C_(a+b-k))`
`=(a)/(a+b)underset(i-0)overset(k)sum(""^(a-1)C_(a-1i)""^(b)C_(k-i))/(""^(a+b-1)C_(a+b-k-1))`
`=(a)/(a+b)underset(i-0)overset(k)sum(""^(a-1)C_(i)""^(b)C_(k-i))/(""^(a+b-1)C_(k))`
`=(a)/(a+b(""^(a+b-1)C_(k)))underset(i-0)overset(k)sum""^(a-1)C_(i)""^(b)C_(k-i)`
`=(a)/(a+b(""^(a+b-1)C_(k)))""^(a+b-1)C_(k)=(a)/(a+b)`