An urn contains m white and n black balls. A ball is drawn at random and is put back into urn along with k additional balls of the same color as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?

Let U={m white, n black balls}

`E_(1)`={First ball drawn of white colour}

`E_(2)`={First ball drawn of black colour}

and `E_(3)`={Second ball of white colour}

`therefore P(E_(1))=m/(m+n)and P(E_(2))=n/(m+n)`

Also, `P(E_(3)//E_(1))=(m+k)/(m+n+k)and P(E_(3)//E_(2))=m/(m+n+k)`

`thereforeP(E_(3))=P(E_(1))cdotP(E_(3)//E_(1))+P(E_(2))cdotP(E_(3)//E_(2))`

`=m/(m+n)cdot(m+k)/(m+n+k)+n/(m+n)cdotm/(m+n+k)`

`=(m(m+k)+nm)/((m+n+k)(m+n))=(m^(2)+mk+nm)/((m+n+k)(m+n))`

`=(m(m+k+m))/((m+n+k)(m+n))=m/(m+n)`

Hence, the probality of drawing a white ball does not depend on k.

Let W

_{1}(B_{1}) be the event that a white (a black) ball is drawn in the first draw and let W be the event that a white ball is drawn in the second draw. ThenP(W) = P(B

_{1}). P(W| B_{1}) + P(W_{1}). P (W| W_{1})= n/m + n. m/m + n + k + m/m + n. m + k/m + n + k

= m(n + m + k)/(m + n) (m + n + k) = m/m + n