If for a binomial distribution *P* (*X* = 1) = *P* (*X* = 2) = α, write *P* (*X* = 4) in terms of α.

#### Solution

\[\text{ For binomial distribution of X } , \]

\[P(X = r) = ^{n}{}{C}_r (p )^r (q )^{n - r} , r = 0, 1, 2, . . . , n\]

\[P(X = 1) = np(q )^{n - 1} \]

\[P(X = 2) =^{n}{}{C}_2 p^2 (q )^{n - 2} \]

\[ \Rightarrow np(q )^{n - 1} = ^{n}{}{C}_2 p^2 (q )^{n - 2} = \alpha \]

\[\text{ Simplifying the above equation we get,} \]

\[q = \frac{n - 1}{2}p\]

\[ \Rightarrow 2q = np - p \]

\[\text{ On putting, q = 1 - p we get } \]

\[2 - 2p = np - p \]

\[p(n + 1) = 2 . . . . . (i)\]

\[\text{ Also} , P(X = 1) = \alpha\]

\[ \Rightarrow np(1 - p )^{n - 1} = \alpha . . . . . (ii)\]

Note: We cannot find the value of n as (*i*) and (*ii*) are not linear and hence we cannot find the value of *P*(*X* = 4)