Let X denote the number of times heads occur in `n` tosses of a fair coin. If `P(X=4),\ P(X=5)\ a n d\ P(X=6)` are in AP; the value of `n` is

Options

A. `7, 14`
B. `10 , 14`
C. `12 ,7`
D. `14 , 12`

Answer

A, D

Explanation

`(a,d)` From the given condition, it follows that `"^(n)C_(4),^(n)C_(5)` and `.^(n)C_(6)` are in `A.P.`
i.e., `2.^(n)C_(5)=^(n)C_(4)+^(n)C_(6)`
`(2n(n-1)(n-2)(n-3)(n-4))/(5!)`
`=(n(n-1)(n-2)(n-3))/(4!)+(n(n-1)(n-2)(n-3)(n-4)(n-5))/(6!)`
`implies(2(n-4))/(5)=1+((n-4)(n-5))/(5.6)`
`impliesn^(2)-21n+98=0`
`implies(n-7)(n-14)=0`
`impliesn=7`, `n=14`