An urn contains 2 white and 2 black balls .A ball is drawn at random.If it is white ,it is not replace into urn ,otherwise it is replaced along with another ball of the same colour .the process is repeated , then find the probabiltiy that the third ball drawn is black.

Answer

B, C

Explanation

Let `B_(i)=` ith ball drawn is black.
`W_(i)=` ith ball drawn is white, where I = 1,2
and A= third ball drawn is black.
We observe that the black ball can be drawn in the third drawn in one of the following mutually exclusive ways.
(i) both first and second balls drawn are white and third ball drawn is black.
i.e. `(W_(1)nnW_(2))nnA`
(ii) Both first and second balls are black and third ball drawn is black.
i.e. `(B_(1)nnB _(2))nnA`
(iii) The first ball drawn is white, the second ball drawn is black and the third ball drawn is black.
i.e. `(W_(1)nnB_(2))nnA`
(iv) The first ball drawn is black, the second ball drawn is white and the third ball drawn is back.
i.e. `(B_(1)nnW_(2))nnA`
`therefore P(A)=P[{(W_(1)nnW_(2))nnA}uu(B_(1)nnB_(2))nnA}`
`uu{(W_(1)nnB_(2))nnA}uu{(B_(1)nnW_(2))nnA}]`
`=P{(W_(1)nnW_(2))nnA}+P{(B_(1)nnB_(2))nnA}`
`+P(W_(1)NNB_(2))nnA}+P{(B_(1)nnW_(2))nnA} `
`=P(W_(1)nnW_(2))*P(A//(W_(1)nnW_(2)))+P(B_(1)nnB_(2))`
`therefore P(A//(B_(1)nnB_(2)))+P(W_(1)nnB_(2))*PP(A//(W_(1)nnB_(2)))`
`+P(B_(1)nnW_(2))*P(A//(B_(1)nnW_(2)))`
`=((2)/(4)xx(1)/(3))xx1+((2)/(4)xx(3)/(5))xx(4)/(5)`
`+((2)/(4)xx(2)/(3))xx(3)/(4)+((2)/(4)xx(2)/(5))xx(3)/(4)`
`=(1)/(6)+(1)/(5)+(1)/(4)+(3)/(20)=(23)/(30)`