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 $P(X \geq 2)$ is:

For a Binomial Distribution with parameters $n$ and $p$:

Mean $= np$

Variance $= npq$, where $q = 1 - p$

We are given:

$(\text{Mean})^2 + (\text{Variance})^2 = \dfrac{225}{256} \quad \cdots(1)$

$(\text{Mean})^2 - (\text{Variance})^2 = \dfrac{63}{256} \quad \cdots(2)$

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Adding $(1)$ and $(2)$:

$2(np)^2 = \dfrac{225}{256} + \dfrac{63}{256} = \dfrac{288}{256}$

$(np)^2 = \dfrac{144}{256}$

$np = \dfrac{12}{16} = \dfrac{3}{4}$

Subtracting $(2)$ from $(1)$:

$2(npq)^2 = \dfrac{225}{256} - \dfrac{63}{256} = \dfrac{162}{256}$

$(npq)^2 = \dfrac{81}{256}$

$npq = \dfrac{9}{16}$

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$q = \dfrac{npq}{np} = \dfrac{9/16}{3/4} = \dfrac{9}{16} \times \dfrac{4}{3} = \dfrac{3}{4}$

$p = 1 - q = 1 - \dfrac{3}{4} = \dfrac{1}{4}$

From $np = \dfrac{3}{4}$:

$n \times \dfrac{1}{4} = \dfrac{3}{4}$

$n = 3$

So $n = 3$, $p = \dfrac{1}{4}$, $q = \dfrac{3}{4}$

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

$P(X=2) = \binom{3}{2}\left(\dfrac{1}{4}\right)^2\left(\dfrac{3}{4}\right)^1 = 3 \times \dfrac{1}{16} \times \dfrac{3}{4} = \dfrac{9}{64}$

$P(X=3) = \binom{3}{3}\left(\dfrac{1}{4}\right)^3 = 1 \times \dfrac{1}{64} = \dfrac{1}{64}$

$P(X \geq 2) = \dfrac{9}{64} + \dfrac{1}{64} = \dfrac{10}{64}$

$P(X \geq 2) = \dfrac{5}{32}$