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.