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Question

(A) The total number of ways to draw $n$ cards from $52$ is $52 	imes 51 	imes \dots 	imes (52 - n + 1)$.For the $n^{th}$ card to be the second ace

Vedclass · Accepted Answer

$\frac{(n - 1)(52 - n)(51 - n)}{50 	imes 49 	imes 17 	imes 13}$

Vedclass · Answer

(A) The total number of ways to draw $n$ cards from $52$ is $52 	imes 51 	imes \dots 	imes (52 - n + 1)$.For the $n^{th}$ card to be the second ace,we must have exactly one ace in the first $(n-1)$ cards and an ace on the $n^{th}$ draw.The number of ways to choose the position of the first ace in the first $(n-1)$ draws is $(n-1)$.The number of ways to choose the first ace is $4$,and the remaining $(n-2)$ cards are chosen from the $48$ non-ace cards in $P(48, n-2)$ ways.The number of ways to choose the second ace on the $n^{th}$ draw is $3$.The total number of favorable outcomes is $(n-1) 	imes 4 	imes P(48, n-2) 	imes 3$.The total number of outcomes for $n$ draws is $P(52, n)$.$P(N=n) = \frac{(n-1) 	imes 4 	imes 3 	imes \frac{48!}{(48-(n-2))!}}{\frac{52!}{(52-n)!}} = \frac{12(n-1) 	imes 48! 	imes (52-n)!}{52! 	imes (50-n)!} = \frac{12(n-1)(52-n)(51-n)}{52 	imes 51 	imes 50 	imes 49} = \frac{(n-1)(52-n)(51-n)}{50 	imes 49 	imes 17 	imes 13}$.

Cards are drawn one by one at random from a well-shuffled full pack of $52$ cards until two aces are obtained for the first time. If $N$ is the number of cards required to be drawn,then $P(N = n)$,where $2 \le n \le 50$,is

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