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Cards are drawn one by one at random from a well suffied full pack of 52 playing card until 2 aces are obtained for the first time. Then prove that probability that exactly n cards are drawn, is `(n-1(52-n)(51-n))/(50xx49xx17xx13),where 2ltnle50.`

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The correct Answer is:
A
P(Nth draw gives 2nd ace)
`=P{"1 ace and "(n-2)" other cards are drawn in "(N-1)" draws")xxP{Nth" draw is 2nd ace"}`
`=(4.(48)!.(n-1)!(52-n)!)/((52)!.(n-2)!(50-n)!).(3)/((58-n))`
`=(4(n-1)(52-n)(51-n).3)/(52.51.50.49)`
`=((n-1)(52-n)(51-n))/(50.49.17.13)`
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