A) \[\frac{(n-1)\,(52-n)\,(51-n)}{50\times 49\times 17\times 13}\]
B) \[\frac{2\,(n-1)\,(52-n)\,(51-n)}{50\times 49\times 17\times 13}\]
C) \[\frac{3\,(n-1)\,(52-n)\,(51-n)}{50\times 49\times 17\times 13}\]
D) \[\frac{4(n-1)\,(52-n)\,(51-n)}{50\times 49\times 17\times 13}\]
Correct Answer: A
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 \[{{n}^{th}}\] draw we get one ace. Hence the required probability \[=\frac{{}^{4}{{C}_{1}}\times {}^{48}{{C}_{n-2}}}{{}^{52}{{C}_{n-1}}}\times \frac{3}{52-(n-1)}\] \[=\frac{4\times (48)\,!}{(n-2)\,!\,(48-n+2)\,!}\times \frac{(n-1)\,!\,(52-n+1)\,!}{(52)\,!}\times \frac{3}{52-n+1}\] \[=\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 13}\] (on simplification).
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