Answer
(a)
$$
E=a_{1} p_{1}+a_{2} p_{2}+\cdots+a_{n} p_{n}
$$
---
(b)
$$
E \approx-1.08
$$
Work Step by Step
(a)
the expected value of this game will be
$$
E=a_{1} p_{1}+a_{2} p_{2}+\cdots+a_{n} p_{n}
$$
(b)
the expected value is:
$$
E=10\left(\frac{4}{52}\right)-2\left(\frac{48}{52}\right) \approx-1.08
$$
