Jitender Pal
Last Activity: 10 Years ago
Hello Student,
Please find the answer to your question
We must have one ace in (n – 1) attempts and one ace in the nth attempt. The probability of drawing one ace in first
(n – 1) attempts is 4C1 x 48Cn – 2/52Cn -1 and other one ace in the nth attempt is, 3C1/[52 – (n – 1)] = 3/53 - n
Hence the required probability,
= 4.48!/(n – 2)! (50 – n)! x (n – 1)! (53 – n)/52! x 3/53 - n
= (n – 1) (52 – n) (51 – n)/50. 49. 17. 13
Thanks
Jitender Pal
askIITians Faculty