Mutually exclusive events reward Say I have an unbiased coin and if I roll heads I get 40 pennies reward and If I roll tails I get 80 pennies. I believe the following is correct for the expected reward from one toss:
reward = p(heads)*40 + p(tails)*80

however is there a formal proof?
 A: You are looking for LOTUS - or the Law of the unconscious statistician: 
https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician
Your Bernoulli random variable $X$ has two values, 1, if tails (with probability $p$), and 0 otherwise. You can find the expectation of the Bernoulli random variable in a standard way: 
$\quad\quad X \sim Bernoulli(p) \\ \quad\quad E[X] = p * 1 + (1 - p) * 0$
Now, define a discrete function $f(x)$:
$ \quad\quad f(x) = \begin{cases}
      80, & \text{if}\ x=1 \\
      40, & \text{otherwise}
    \end{cases}$ 
$f(X)$ is also a random variable (function of a random variable is random variable), which has 2 outcomes: 40 and 80. This is exactly what you need since $Y = f(X)$ is actually a random variable that denotes the amount of money you win from your bet. Now, using LOTUS: 
$\quad\quad E[Y] = E[f(X)] = \sum_{i \in \{0, 1\}} f(i)P(X = i) = 80 * p + 40 * (1 - p)$
IF it is a fair coin with $p = 0.5$, you have:
$\quad\quad E[f(X)] = 80 * 0.5 + 40 * 0.5 = 60$
Q.E.D. that your solution was right
