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Algebra

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Joined:	11 Jun 2007
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11 Jun 2007 16:56:49 IST

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From permutations and combinations

None

1.Cards are drawn one by 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 be drawn,( 2

50) then show that

P {N=n} = (n-1)(52-n)(51-n)

50x49x17x13

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PRIYAM DAS

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Joined: 9 Jun 2007

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11 Jun 2007 22:53:23 IST

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INTELLIGENT

Amar Gupta

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Joined: 31 Jan 2007

Posts: 363

13 Jun 2007 09:31:37 IST

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Dear,

let the game is over at nth card. so this must be second ace and the first ace must come in the previous (n-1) cards.

P(n) = prob. of getting ace in first (n-1) cards * prob. of getting ace in the nth card

prob. of getting ace in first (n-1) cards = no. of favourable cases(M) / total possible cases(N)

[ there are 4 aces so no. of ways of getting one ace =4C1
no. of ways to get rest (n-2) cards = 48C(n-2)
now we can arrange them in (n-1)! ways]

so M = 4C1 * 48C(n-2) *(n-1)!

and N = 52C(n-1) * (n-1)!

so prob. of getting ace in first (n-1) cards = 4C1 * 48C(n-2) / 52C(n-1)

or = [4 * 48! * (n-1)! * (53-n)!] / [ (n-2)! * (50-n)! * 52! ]

or = [ 4 (n-1) (53-n) (52-n) (51-n) ] / 52*51*50*49

and hence P(n) = [ {4 (n-1) (53-n) (52-n) (51-n) } / (52*51*50*49) ] * [3/(53-n)]

where 3/(53-n) = prob. of getting ace at nth card

so P(n) =12 (n-1) (52-n) (51-n) / (52*51*50*49)

or P(n) = (n-1)(52-n)(51-n) / (13*17*50*49)

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