Solveeit Logo
Login

Question

Question: Cards are drawn one by one at random from a well shuffled full pack of 52 cards until two aces are o...

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 Pr {N = n}, where 2 £ n £ 50, is –

A

(n−1)(52−n)(51−n)50×49×17×13\frac { ( \mathrm { n } - 1 ) ( 52 - \mathrm { n } ) ( 51 - \mathrm { n } ) } { 50 \times 49 \times 17 \times 13 }

B

2(n−1)(52−n)(51−n)50×49×17×13\frac { 2 ( \mathrm { n } - 1 ) ( 52 - \mathrm { n } ) ( 51 - \mathrm { n } ) } { 50 \times 49 \times 17 \times 13 }

C

3(n−1)(52−n)(51−n)50×49×17×13\frac { 3 ( n - 1 ) ( 52 - n ) ( 51 - n ) } { 50 \times 49 \times 17 \times 13 }

D

4(n−1)(52−n)(51−n)50×49×17×13\frac { 4 ( n - 1 ) ( 52 - n ) ( 51 - n ) } { 50 \times 49 \times 17 \times 13 }

Answer

(n−1)(52−n)(51−n)50×49×17×13\frac { ( \mathrm { n } - 1 ) ( 52 - \mathrm { n } ) ( 51 - \mathrm { n } ) } { 50 \times 49 \times 17 \times 13 }

Explanation

Solution

Here the least number of draws to obtain 2 aces are 2 and the maximum number is 50 thus n can take value from 2 to 50. Since we have to make n draws for getting two aces, in

(n – 1) draws, we get any one of the 4 aces and in the nth draw we get one ace. Hence the required probability

= 4C1×48Cn−252Cn−1\frac { { } ^ { 4 } C _ { 1 } \times { } ^ { 48 } C _ { n - 2 } } { { } ^ { 52 } C _ { n - 1 } } 352−(n−1)\frac { 3 } { 52 - ( n - 1 ) }

= 4×(48)!(n−2)!(48−n+2)!\frac { 4 \times ( 48 ) ! } { ( n - 2 ) ! ( 48 - n + 2 ) ! }× (n−1)!(52−n+1)!(52)!\frac { ( \mathrm { n } - 1 ) ! ( 52 - \mathrm { n } + 1 ) ! } { ( 52 ) ! } 352−n+1\frac { 3 } { 52 - n + 1 }

= (n−1)(52−n)(51−n)50×49×17×13\frac { ( \mathrm { n } - 1 ) ( 52 - \mathrm { n } ) ( 51 - \mathrm { n } ) } { 50 \times 49 \times 17 \times 13 } (on simplification)