Question

If the mean and variance of a binomial distribution are $\frac{4}{3}$ and $\frac{10}{9}$ respectively, then $P(X \geq 6) =$

• A. $41/6^8$
• B. $741/6^8$
• C. $1-741/6^8$
• D. $1-41/6^8$

Hint:

For a binomial distribution, you can quickly find the parameters by remembering that $q = \frac{\text{Variance}}{\text{Mean}}$ and $p=1-q$. Once you have $p$, you can find $n$ from the mean formula, $n = \text{Mean}/p$.

Answer 1

The Correct Option is B

For a binomial distribution $B(n, p)$, the mean is $\mu = np$ and the variance is $\sigma^2 = npq$, where $q = 1-p$.
We are given:
Mean: $np = \frac{4}{3}$.
Variance: $npq = \frac{10}{9}$.
Step 1: Find the parameters $n, p, q$.
Divide the variance by the mean to find $q$:
$q = \frac{npq}{np} = \frac{10/9}{4/3} = \frac{10}{9} \times \frac{3}{4} = \frac{10}{12} = \frac{5}{6}$.
Now find $p$:
$p = 1 - q = 1 - \frac{5}{6} = \frac{1}{6}$.
Now find $n$ using the mean:
$n \cdot p = \frac{4}{3} \implies n \cdot \frac{1}{6} = \frac{4}{3} \implies n = \frac{4}{3} \times 6 = 8$.
So, the distribution is $X \sim B(n=8, p=1/6)$.
Step 2: Calculate $P(X \geq 6)$.
This is equal to $P(X=6) + P(X=7) + P(X=8)$.
The probability mass function is $P(X=k) = \binom{n}{k} p^k q^{n-k}$.
$P(X=6) = \binom{8}{6} (\frac{1}{6})^6 (\frac{5}{6})^{8-6} = \frac{8 \cdot 7}{2} \cdot (\frac{1}{6})^6 (\frac{5}{6})^2 = 28 \cdot \frac{25}{6^8} = \frac{700}{6^8}$.
$P(X=7) = \binom{8}{7} (\frac{1}{6})^7 (\frac{5}{6})^{8-7} = 8 \cdot (\frac{1}{6})^7 (\frac{5}{6})^1 = \frac{40}{6^8}$.
$P(X=8) = \binom{8}{8} (\frac{1}{6})^8 (\frac{5}{6})^{8-8} = 1 \cdot (\frac{1}{6})^8 (1) = \frac{1}{6^8}$.
$P(X \geq 6) = \frac{700}{6^8} + \frac{40}{6^8} + \frac{1}{6^8} = \frac{700+40+1}{6^8} = \frac{741}{6^8}$.