Cards are drawn 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 the drawn, then show that `P ,{N=n}=((n-1)(52-n)(51-n))/(50xx49xx17xx13)`, where

Answer

A

Explanation

P(Nth draw gives 2nd ace)
`=P{"1 ace and "(n-2)" other cards are drawn in "(N-1)" draws")xxP{Nth" draw is 2nd ace"}`
`=(4.(48)!.(n-1)!(52-n)!)/((52)!.(n-2)!(50-n)!).(3)/((58-n))`
`=(4(n-1)(52-n)(51-n).3)/(52.51.50.49)`
`=((n-1)(52-n)(51-n))/(50.49.17.13)`