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Question

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replaced into urn, otherwise it is replaced along with another ball of the same colour. The process is repeated.The probability that the third ball drawn is black is 23×k60.then the value of k is

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Solution

There are four possibilities
First ballSecond ballEvent
White WhiteE1
White BlackE2
BlackWhiteE3
BlackBlackE4
P(E1)=24×13=16
P(E2)=24×23=13
P(E3)=24×25=15
P(E4)=24×35=310
Let E denotes the event of drawing a black ball in the third attempt
P(EE1)=1 because there is no white ball left to be selected.
P(EE2)=34 because there are 3 black ball and 1 white ball left.
P(EE3)=34 because again there are 3 black and 1 white ball left.
P(EE4)=46=23 because there are 4 black balls and 2 white balls.
P(E)=P(E1)P(EE1)+P(E2)P(EE2)+P(E3)P(EE3)+P(E4)P(EE4)
=16×1+13×34+15×34+310×23=16+14+320+15=2330k=2

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