= Solution
The three preceding <probabilities> satisfy
$$
\boxed{\frac{28}{45}+\frac{16}{45}+\frac1{45}=1.}
$$
Their <expected value> and second <moment> are
$$
\mu=\frac{16+2}{45}=\frac25,\qquad
\mathbb E[X^2]=\frac{16+4}{45}=\frac49.
$$
Subtracting the squared <mean> gives
$$
\boxed{\mu=\frac25,\qquad\sigma^2=\frac49-\frac4{25}=\frac{64}{225}.}
$$
The value $\mu=2(2/10)$ also follows from <linearity of expectation>. The <variance> includes the finite-population reduction caused by <sampling without replacement>.
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