If the binomial distribution $X\sim B(n, p)$ of mean 3 and variance $\frac{3}{2}$, $(p + q) = 1$, then which of the following is/are TRUE?

(A) $q = \frac{1}{2}$, $n = 6$

(B) $P(X \leq 5) = \frac{63}{64}$, $p = \frac{1}{2}$

(C) $q = \frac{1}{3}$, $p = \frac{2}{3}$

(D) $P(X = 4) = \frac{15}{64}$, $n = 6$

Choose the correct answer from the options given below:

For a binomial distribution $X \sim B(n,p)$:

Mean $= np = 3$

Variance $= npq = \frac{3}{2}$

Given $p + q = 1$

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Dividing variance by mean:

$\frac{npq}{np} = \frac{3/2}{3}$

$q = \frac{1}{2}$

Since $p + q = 1$:

$p = 1 - \frac{1}{2}$

$p = \frac{1}{2}$

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Using $np = 3$ and $p = \frac{1}{2}$:

$n \cdot \frac{1}{2} = 3$

$n = 6$

Therefore: $n = 6$, $p = \frac{1}{2}$, $q = \frac{1}{2}$

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Checking option (A): $q = \frac{1}{2}$, $n = 6$

This matches our values. Option (A) is TRUE.

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Checking option (B): $P(X \leq 5) =\frac{63}{64}$, $p = \frac{1}{2}$

$P(X \leq 5) = 1 - P(X = 6)$

$P(X \leq 5) = 1 - \binom{6}{6}\left(\frac{1}{2}\right)^6$

$P(X \leq 5) = 1 - \frac{1}{64} = \frac{63}{64}$

Since $P(X \leq 5)= \frac{63}{64}$, option (B) is TRUE.

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Checking option (C): $q = \frac{1}{3}$, $p = \frac{2}{3}$

This contradicts our calculated values. Option (C) is FALSE.

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Checking option (D): $P(X = 4) = \frac{15}{64}$, $n = 6$

$P(X = 4) = \binom{6}{4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^2$

$P(X = 4) = 15 \times \frac{1}{64}$

$P(X = 4) = \frac{15}{64}$

Option (D) is TRUE.

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Therefore, options (A), (B) and (D) are correct.