On the first roll, you have a 1/3 probability of rolling a 1 or 2 and thus winning $200.
There's a 2/3 probability of not rolling a 1 or 2 on the first roll. On the second roll, there is again a 1/3 probability of rolling a 1 or 2, and 2/3 probability otherwise, so that there is a 1/3*2/3 = 2/9 probability of getting a 1 or 2 and thus winning $100, and 2/3*2/3 = 4/9 probability of losing.
(a) Let [tex]W[/tex] be a random variable representing the winnings from playing the game. It has the probability mass function
[tex]P(W=w)=\begin{cases}\frac13&\text{for }w=\$200\\\frac29&\text{for }w=\$100\\\frac49&\text{otherwise}\end{cases}[/tex]
(b) Compute the expected value of [tex]W[/tex]:
[tex]E[W]=\displaystyle\sum_w w\,P(W=w)=\$200\cdot\frac13+\$100\cdot\frac29+\$0\cdot\frac49\approx\$88.89[/tex]
