Question

If $X\sim B(n,p)$ is a binomial variate, $8p^2+15p-2=0$ and the product of the mean and variance of $X$ is $\dfrac78$, then $P(X=4)=$

• A. ${}^{6}C_{4}\dfrac{5^2}{6^6}$
• B. ${}^{7}C_{4}\dfrac{4^3}{5^7}$
• C. ${}^{5}C_{4}\dfrac{3}{4^5}$
• D. ${}^{8}C_{4}\dfrac{7^4}{8^8}$

Hint:

For a binomial variable, remember: Mean $=np$ and Variance $=npq$.

Answer 1

The Correct Option is D

Step 1: Solve for $p$.
We are given $8p^2+15p-2=0.$ Factor: $(8p-1)(p+2)=0$, so $p=\dfrac18$ or $p=-2.$ A probability must be between $0$ and $1$, so $p=\dfrac18.$
Step 2: Find $q$.
$q=1-p=1-\dfrac18=\dfrac78.$
Step 3: Recall mean and variance.
For $B(n,p)$ the mean is $np$ and the variance is $npq.$
Step 4: Use the product condition.
Given $(np)(npq)=\dfrac78$, that is $n^2p^2q=\dfrac78.$ Put $p^2=\dfrac1{64}$ and $q=\dfrac78$: \[ n^2\cdot\frac1{64}\cdot\frac78=\frac78. \]
Step 5: Solve for $n$.
Cancel $\dfrac78$ on both sides: $\dfrac{n^2}{64}=1$, so $n^2=64$ and $n=8.$
Step 6: Compute $P(X=4)$.
$P(X=4)=\binom84 p^4 q^4=\binom84\left(\dfrac18\right)^4\left(\dfrac78\right)^4=\binom84\dfrac{7^4}{8^8}.$ \[ \boxed{{}^{8}C_{4}\dfrac{7^4}{8^8}} \]