A fair coin is tosed `n` times. Let `X` denote the number of times head occurs. If `P(X=4)`, `P(X=5)` and `P(X=6)` are in arithmetic progression, then the value of `n` can be

Options

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

Answer

A

Explanation

Clearly, X is a binomial variate with parameters n and `p=1//2` such that
`P(X=r)=.^n C_rp^rq^n-1=.^nC_r((1)/(2))^r((1)/(2))^(n-r)=.^nC_r((1)/(2))^r` ltNow,
`P(X=4),P(X=5)and P(X=6)`are in A.P.
`rArr 2P(X=5)=P(X=4)+P(X=6)`
`2.^nC_5((1)/(2))^n=.^nC_4((1)/(2))^n+.^nC_6((1)/(2))^n`
`rArr 2.^nC_5=.^C_4+.^nC_6`
`rArr 2(ǃ)/((n-5)ǃ5ǃ)=(nǃ)/((n-4)ǃ4ǃ)+(nǃ)/((n-6)ǃ6ǃ)`
`rArr (2)/(5(n-5))=(1)/((n-4)(n-5))+(nǃ)/(6x5)`
`rArr n^2-21n+98=0rArr (n-7)(n-14)=0rArr n=7or 14`.