Answer
$E(prize)=0.58$
$E(loss)=0.42$
It is expected that a player will lose 0.42 per ticket bought.
If a player buys 1000 tickets, it is expected that he will lose $420.
Work Step by Step
$E(prize)=E(X)=μ_X=Σ[x.P(x)]=100,000\times\frac{1}{324,632}+300\times\frac{1}{2164}+10\times\frac{1}{75}+0\times P(0)=\frac{100,000}{324,632}+\frac{300}{2164}+\frac{10}{75}+0=0.58$
Notice that we could have computed the value of P(0). But, since $0\times P(0)=0$, it was not necessary.
The expected profit is the difference between the expected prize and the price of a single ticket.
$E(profit)=0.58-1=-0.42$
or
$E(loss)=0.42$
