A gambler plays seven games one after the other and the chance to win each of them is $\frac{1}{3}$, independently of the others. For $k = 1, ..., 7$, if the gambler wins game number $k$, then the gambler gets $k$ gold coins. Denote by $X$ the total number of gold coins won in all seven games.
$a$. Compute $P(X = 3)$
$b$. Compute $E(X)$
For part $a$ I know the only way to get $3$ coins is if the gambler only wins games $1$ and $2$ or only wins game $3$. I'm not sure how to get the actual probability for this. Is there a formula I should use?
For part $b$ is it just $E(X) = 1(\frac{1}{3}) + 2(\frac{1}{3}) + 3(\frac{1}{3}) + ... + 7(\frac{1}{3}) $?
Your reasoning for a is correct. How do you calculate the probability of the outcome that he wins a specified number of games out of the seven. Since the games are independent, the probability of him winning game $i$ is independent of the other games. Let $W_i$ denote the event that he wins game $i$ and $W_i^c$ the event that he loses game $i$. Thus the event that he wins the first two games and loses the others is $$P(W_1\cap W_2\cap W_3^c \cap W_4^c \cap W_5^c \cap W_6^c \cap W_7^c)=P(W_1)P(W_2)P(W_3^c)P(W_4^c)P(W_5^c)P(W_6^c)P(W_7^c)$$ by independence. And you know how to calculate each of the probabilities on the right.