Question

Let X denote the number of times heads occur in n tosses of a fair coin. If P (X = 4), P (X = 5) and P (X = 6) are in AP, the value of n is
(a) 7, 14
(b) 10, 14
(c) 12, 7
(d) 14, 12

Answer 1

(a) 7, 14

$$\begin{gathered} \mathrm{Here},p=\frac{1}{2}\mathrm{and}q=\frac{1}{2} \\ \mathrm{Binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=r)={}^{n}C_r{\left(\frac{1}{2}\right)}^r{\left(\frac{1}{2}\right)}^{n-r} \end{gathered}$$

P (X = 4), P (X = 5), P(X = 6) are in A.P.

$$\begin{gathered} \therefore {}^{n}C_4+{}^{n}C_6=2{}^{n}C_5 \\ \Rightarrow \frac{n(n-1)(n-2)(n-3)}{2\left(4!\right)}+\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{2\left(6!\right)}=\frac{n(n-1)(n-2)(n-3)(n-4)}{5!} \\ \mathrm{By}\mathrm{simplifying},\mathrm{we}\text{get} \\ \frac{1}{2}+\frac{(n-4)(n-5)}{2\left(30\right)}=\frac{n-4}{5} \\ \mathrm{Taking}\mathrm{LCM}\hspace{0.17em}\mathrm{as}60,\text{we get} \\ 30+n^2-9n+20=12\mathrm{n}-48 \\ \Rightarrow n^2-21n+98=0 \\ \Rightarrow (n-7)(n-14)=0 \\ \Rightarrow n=7,14 \end{gathered}$$