MCQ
If $X$ follows a binomial distribution with parameters $n = 6$ and $p$ and $4\,(P(X = 4)) = P(X = 2),$ then $p = $
- A$\frac{1}{2}$
- B$\frac{1}{4}$
- C$\frac{1}{6}$
- ✓$\frac{1}{3}$
$ \Rightarrow 4{p^2} = {q^2} \Rightarrow 4{p^2} = {(1 - p)^2}$
$ \Rightarrow 3{p^2} + 2p - 1 = 0 \Rightarrow p = \frac{1}{3}$.
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