1. **Problem statement:** Given a binomial distribution with mean $\mu = 4$ and variance $\sigma^2 = \frac{4}{3}$, find the probability $P(X > 1)$. 2. **Recall formulas for binomial distribution:** Mean: $\mu = np$ Variance: $\sigma^2 = np(1-p)$ where $n$ is the number of trials and $p$ is the probability of success in each trial. 3. **Use the given mean and variance to find $n$ and $p$: From mean: $np = 4$ From variance: $np(1-p) = \frac{4}{3}$ 4. **Substitute $np=4$ into variance equation:** $4(1-p) = \frac{4}{3}$ Divide both sides by 4: $1-p = \frac{1}{3}$ So, $p = 1 - \frac{1}{3} = \frac{2}{3}$ 5. **Find $n$ using $np=4$: $n \times \frac{2}{3} = 4$ $n = \frac{4 \times 3}{2} = 6$ 6. **Now we have $n=6$ and $p=\frac{2}{3}$. The binomial distribution is $X \sim Binomial(6, \frac{2}{3})$. 7. **Calculate $P(X > 1)$:** $P(X > 1) = 1 - P(X \leq 1) = 1 - [P(X=0) + P(X=1)]$ 8. **Calculate $P(X=0)$ and $P(X=1)$ using binomial formula:** $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - $P(X=0) = \binom{6}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^6 = 1 \times 1 \times \left(\frac{1}{3}\right)^6 = \frac{1}{729}$ - $P(X=1) = \binom{6}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^5 = 6 \times \frac{2}{3} \times \frac{1}{243} = 6 \times \frac{2}{3} \times \frac{1}{243} = \frac{12}{729}$ 9. **Sum these probabilities:** $P(X \leq 1) = \frac{1}{729} + \frac{12}{729} = \frac{13}{729}$ 10. **Finally, calculate $P(X > 1)$:** $$P(X > 1) = 1 - \frac{13}{729} = \frac{716}{729} \approx 0.9811$$ **Answer:** $P(X > 1) = \frac{716}{729} \approx 0.9811$