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Cards are drawn one at random from a well shuffled full pack of 52 playing cards until 2 aces are obtained for the first time. If `N` is the number of cards required to the drawn, then show that `P ,{N=n}=((n-1)(52-n)(51-n))/(50xx49xx17xx13),w h e r e2

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We must have one ace in `(n-1)` attempts and one ace in the nth attempt.
Let event A be one ace card and `(n-2)` other than ace cards are drawn in first `(n-1)` draws.
`thereforeP(A)=(""^(4)C_(1)xx""^(48)C_(n-2))/(""^(52)C_(n-1))`
Let event B be nth card is ace.
Therefore, required probability is
`P(AnnB)=P(A)P(B//A)`
where `P(B//A)=(""^(3)C_(1))/(""^([52-(n-1)])C_(1))=(3)/(53-n)`
`thereforeP(AnnB)=(4xx48!)/((n-2)(52-n)(51-n))xx((n-)!(53-n)!)/(52!)xx(3)/(53-n)`
`=((n-1)(52-n)(51-n))/(50xx49xx17xx13)`
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