Probability
Binomial Distribution — Mean-Variance and Finding n²P(X>1)
nta_pyq_2023_apr
Grade 12
Question:
$X\sim B(n,p)$, mean $-$ variance $=1$, $2P(X=2)=3P(X=1)$. Then $n^2P(X>1)$ is equal to
Step-by-Step Solution
Key Concept: $np-npq=np(1-q)=np^2=1$ and $\frac{2\binom{n}{2}p^2q^{n-2}}{\binom{n}{1}pq^{n-1}}=\frac{3}{1}\Rightarrow(n-1)p=3q$.
$n=4,p=\frac{1}{2}$. $n^2P(X>1)=11$.
Correct Answer: 2