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Question

From a bag containing n balls, all either white or black, all numbers of each being equally likely, a ball is drawn which turns out to be white; this is replaced, and another ball is drawn, which also turns out to be white. If this ball is replaced, prove that the chance of the next draw giving a black ball is 12(n1)(2n+1)1.

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Solution

We have n cases to consider, for there may be 1,2,3,...n white balls; and all these cases are equally likely,
so that
P1=P2=P3=.....=Pn
If there were r white balls, the chance of drawing two white balls in this case should be (rn)2
Q1(1n)2=Q2(2n)2=....Qr(rn)2=.....=1r2n2
Thus Qr(rn)2=6n(n+1)(2n+1),Qr=6r2n(n+1)(2n+1)
And the chance of another drawung giving a black ball
r=nr=1nrn.6r2n(n+1)(2n+1)=6r2n(n+1)(2n+1)6r3n2(n+1)(2n+1)
=16n2(n+1)24n2(n+1)(2n+1)=13(n+1)2(2n+1)=12(n1)(2n+1)1

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