If the binomial distribution \(X \sim B(n,p)\) of mean 3 and variance \(\displaystyle\frac{3}{2}\), \((p+q) = 1\), then which of the following is/are TRUE?
(A) \(q=\displaystyle\frac{1}{2},n=6\)
(B) \(P(X \le 5)=\displaystyle\frac{63}{64},p=\frac{1}{2}\)
(C) \(q=\displaystyle\frac{1}{3},p=\frac{2}{3}\)
(D) \(P(X = 4) =\displaystyle\frac{15}{64},n=6\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
For a binomial distribution, mean \(= np\) and variance \(= npq\).
\(np = 3\)
\( \displaystyle npq = \frac{3}{2} \)
Dividing the second by the first:
\( \displaystyle q = \frac{1}{2} \)
\( \displaystyle p = 1 - q = \frac{1}{2} \)
\( \displaystyle n = \frac{3}{p} = 6 \)
With \(n = 6\) and \(p = q = \frac{1}{2}\), every probability has the denominator \(2^6 = 64\):
\( \displaystyle P(X \le 5) = 1 - P(X = 6) = 1 - \frac{1}{64} = \frac{63}{64} \)
\( \displaystyle P(X = 4) = {}^{6}C_{4}\left(\frac{1}{2}\right)^6 = \frac{15}{64} \)
So \(n = 6\), \(p = q = \frac{1}{2}\), \(P(X \le 5) = \frac{63}{64}\) and \(P(X = 4) = \frac{15}{64}\).
Checking the statements
(A) \(q=\frac{1}{2},n=6\) — Correct. \(q = \frac{1}{2}\) and \(n = 6\).
(B) \(P(X \le 5)=\frac{63}{64},p=\frac{1}{2}\) — Correct. \(1 - \frac{1}{64} = \frac{63}{64}\) and \(p = \frac{1}{2}\).
(C) \(q=\frac{1}{3},p=\frac{2}{3}\) — Incorrect. \(p = q = \frac{1}{2}\).
(D) \(P(X = 4) =\frac{15}{64},n=6\) — Correct. \({}^{6}C_{4} = 15\), so \(P(X = 4) = \frac{15}{64}\), with \(n = 6\).