If the binomial distribution $X~β(n,p)$ 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) $P(X ≤5)=\frac{63}{64},p=\frac{1}{2}$
(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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (4) → (A), (B) and (D) only **
Given $X\sim\mathrm{Bin}(n,p)$ with $\mathbb{E}(X)=np=3$ and $\mathrm{Var}(X)=np(1-p)=\frac{3}{2}$.
From $np=3\Rightarrow n=\frac{3}{p}$.
Substitute into variance: $np(1-p)=3(1-p)=\frac{3}{2}\Rightarrow 1-p=\frac{1}{2}\Rightarrow p=\frac{1}{2}$. Hence $n=\frac{3}{p}=6$ and $q=1-p=\frac{1}{2}$.
Check options:
(A) $q=\frac{1}{2},\;n=6$ — True.
(B) With $n=6,p=\frac{1}{2}$, $P(X\le5)=1-P(X=6)=1-(\frac{1}{2})^{6}=1-\frac{1}{64}=\frac{63}{64}$ — True.
(C) $q=\frac{1}{3},p=\frac{2}{3}$ — False.
(D) With $n=6,p=\frac{1}{2}$, $P(X=4)=15\cdot\frac{1}{64}=\frac{15}{64}$ — True.
Correct: (A), (B) and (D)