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 $p$, the following relationships hold:
\[\text{Mean} = np = \frac{4}{3}, \text{Variance} = np(1 - p) = \frac{8}{9}\] From the mean, we have $np = \frac{4}{3}$, which implies $p = \frac{4}{3n}$.
Substituting this into the variance equation yields:
\[\frac{4}{3}(1 - \frac{4}{3n}) = \frac{8}{9}\] Simplifying the equation:
\[1 - \frac{4}{3n} = \frac{8}{9}.\frac{3}{4} = \frac{2}{3}\] \[\frac{4}{3n} = 1 - \frac{2}{3} = \frac{1}{3}\] \[n = 4\] Therefore, $p = \frac{4}{3 \times 4} = \frac{1}{3}$.
We now calculate $P(X = 1)$ for a binomial distribution with $n = 4$ and $p = \frac{1}{3}$:
\[P(X = 1) = {4 \choose 1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^3 = 4 \times \frac{1}{3} \times \frac{8}{27} = \frac{32}{81} = \boxed{\frac{32}{81}}\]