Question

If the mean and variance of a binomial distribution are $\dfrac{4}{3}$ and $\dfrac{8}{9}$ respectively, then find $P(X = 1)$.

Hint:

Use mean and variance equations to determine $n$ and $p$ in binomial distributions. Then apply the binomial formula to find required probabilities.

Answer 1

For a binomial distribution with parameters $n$ and $p$, we have:
\[ \text{Mean} = np = \frac{4}{3}, \text{Variance} = np(1 - p) = \frac{8}{9} \]
From mean: $np = \dfrac{4}{3} \Rightarrow p = \dfrac{4}{3n}$
Substitute into variance: \[ np(1 - p) = \frac{8}{9} \Rightarrow \frac{4}{3}(1 - \frac{4}{3n}) = \frac{8}{9} \]
Simplify: \[ 1 - \frac{4}{3n} = \frac{8}{9}.\frac{3}{4} = \frac{2}{3} \Rightarrow \frac{4}{3n} = 1 - \frac{2}{3} = \frac{1}{3} \Rightarrow n = 4 \]
Then $p = \dfrac{4}{3.4} = \dfrac{1}{3}$
Now find $P(X = 1)$ in $B(n = 4, p = \frac{1}{3})$:
\[ P(X = 1) = {4 \choose 1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4.\frac{1}{3}.\frac{8}{27} = \frac{32}{81} = \boxed{\frac{32}{81}} \]