MCQ
The random valuable $X$ follows binomial distribution $B (n, p)$ for which the difference of the mean and the variance is $1$. If $2 P(X=2)=3 P(X=1)$, then $n^2 P(X > 1)$ is equal to $......$.
- A$12$
- B$15$
- ✓$11$
- D$16$
$\Rightarrow np ^2=1$
$2^{ n } C _2 p ^2 q ^{ n-2 }=3^{ n } C _1 p q^{n-1}$
$\Rightarrow n p-p=3 q \quad(\therefore q=1-p)$
$\Rightarrow p =\frac{1}{2}$
Hence $n=4$
$P(x > 1)=1-(p(x=0)+p(x=1)$
$=1-\left({ }^4 C _0\left(\frac{1}{2}\right)^4+{ }^4 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^3\right)=\frac{11}{16}$
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