If the sum and difference of squares of mean and variance of a Binomial distribution is $\frac{225}{256}$ and $\frac{63}{256}$ respectively, the $P(X ≥ 2)$ is:

$\frac{5}{32}$

The correct answer is Option (3) → $\frac{5}{32}$ **

Mean $=np$

Variance $=np(1-p)$

Given:

$\displaystyle (np)^{2} + (np(1-p))^{2} = \frac{225}{256}$

$\displaystyle (np)^{2} - (np(1-p))^{2} = \frac{63}{256}$

Add the equations:

$2(np)^{2} = \frac{225+63}{256} = \frac{288}{256} = \frac{9}{8}$

$(np)^{2} = \frac{9}{16}$

$np = \frac{3}{4}$

Subtract the equations:

$2(np(1-p))^{2} = \frac{225-63}{256} = \frac{162}{256} = \frac{81}{128}$

$(np(1-p))^{2} = \frac{81}{256} \Rightarrow np(1-p) = \frac{9}{16}$

Divide:

$\frac{np(1-p)}{np} = \frac{9/16}{3/4} = \frac{9}{16} \cdot \frac{4}{3} = \frac{3}{4}$

$1-p = \frac{3}{4}$

$p = \frac{1}{4}$

$np = \frac{3}{4} \Rightarrow n\cdot \frac{1}{4} = \frac{3}{4} \Rightarrow n = 3$

So distribution is $X \sim B(3,\frac{1}{4})$.

Required:

$P(X \ge 2) = P(X=2) + P(X=3)$

$P(X=2) = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} = \frac{9}{64}$

$P(X=3) = \left(\frac14\right)^{3} = \frac{1}{64}$

$P(X\ge 2) = \frac{9}{64} + \frac{1}{64} = \frac{10}{64} = \frac{5}{32}$

Final Answer: $\displaystyle \frac{5}{32}$