JEE Mathematics 2023 Question with Solution

Given: $X$ follows binomial distribution $B(n,p)$. The difference between mean and variance is $1$, and

$$2P(X=2)=3P(X=1)$$

Find: $n^2P(X>1)$.

Using binomial probabilities,

$$2\binom{n}{2}p^2(1-p)^{n-2}=3\binom{n}{1}p(1-p)^{n-1}$$

So,

$$2\cdot \frac{n(n-1)}{2}p^2(1-p)^{n-2}=3np(1-p)^{n-1}$$

Cancelling common factors gives

$$(n-1)p=3(1-p)$$

Hence,

$$p(n+2)=3$$

so

$$p=\frac{3}{n+2}$$

For a binomial distribution, mean $=np$ and variance $=np(1-p)$. Their difference is

$$np-np(1-p)=np^2=1$$

Thus,

$$np^2=1$$

Substituting $p=\frac{3}{n+2}$,

$$n\left(\frac{3}{n+2}\right)^2=1$$ $$9n=(n+2)^2$$ $$n^2-5n+4=0$$ $$(n-1)(n-4)=0$$

Since $2P(X=2)$ is involved, we need $n\ge 2$. Therefore,

$$n=4$$

Then,

$$p=\frac{3}{4+2}=\frac{1}{2}$$

Now,

$$P(X>1)=1-P(X=0)-P(X=1)$$

For $n=4, p=\frac{1}{2}$,

$$P(X=0)=\binom{4}{0}\left(\frac{1}{2}\right)^4=\frac{1}{16}$$ $$P(X=1)=\binom{4}{1}\left(\frac{1}{2}\right)^4=\frac{4}{16}$$

Therefore,

$$P(X>1)=1-\frac{1}{16}-\frac{4}{16}=\frac{11}{16}$$

Hence,

$$n^2P(X>1)=4^2\cdot \frac{11}{16}=11$$

Therefore, the correct option is B.

The solution states The Correct Option is D, but the worked calculation concludes

$$n^2P(X>1)=11$$

which matches option B. The algebra in the working supports B, so the final answer is taken as B.