Answer: 1.3

The solution involves using the properties of binomial distribution: 1. For a binomial random variable X ~ Bin(n, p): - \(P(X=0) = (1-p)^n = 0.20\) - \(P(X=1) = np(1-p)^{n-1} = 0.35\) - \(E(X) = np = 1.5\) 2. From \(E(X) = np = 1.5\), we can substitute into \(P(X=1)\): \(1.5(1-p)^{n-1} = 0.35\) \((1-p)^{n-1} = 0.35/1.5 = 0.2333\) 3. We also have \((1-p)^n = 0.20\) from \(P(X=0)\) 4. Dividing these two equations: \(\frac{(1-p)^n}{(1-p)^{n-1}} = \frac{0.20}{0.2333}\) \(1-p = 0.8571\) 5. Now calculate variance: \(Var(X) = np(1-p) = 1.5 \times 0.8571 = 1.2857 \approx 1.3\) Therefore, the variance is approximately 1.3, which corresponds to option D.

Author: Nikitesh Somanthe