Rolling a die and then tossing a coin

Question

A fair 6 sided die is tossed. Let its outcome be denoted by X. Then a fair coin is tossed X times and Y denotes the number of heads in X tosses. Calculate (a)P(Y=4) ; (b)P(X=1). I am getting a different answer when I calculate the probability by writing the sample space, and a different answer when I use the binomial formula to calculate the probabilities. Please help.

Answer 1

You can calculate favorable cases.

Let us focus on $p(Y=4)$, since $p(X=1)$ is trivial. The only possibility for $Y$ to attain such value is for values of X in $\{4,5,6\}$ (we can ignore the other ones).

For $X=4$, all $4$ coin tosses would have to land heads for $Y$ to have value $4$. The probability of this event is $(\frac{1}{2})^4=\frac{1}{16}$. If we take into account the probability $p(x=4)$ then we have $p(Y=4|X=4)\,p(X=4)=\frac{1}{96}$.

For $X=5$, four out of five tosses would have to land heads. 5 possible combinations, each with probability $\frac{1}{32}$. Thus, $\frac{5}{32}$. Taking into account $p(X=5)$, we have the probability $p(Y=4|X=5)\,p(X=5)=\frac{5}{192}$.

Finally, for $X=6$, four out of six tosses need to land heads. 15 possible combinations, each with probability $\frac{1}{64}$. Thus, $\frac{15}{64}$. Taking into account $p(X=6)$, we have the probability $p(Y=4|X=6)\,p(X=6)=\frac{15}{384}$.

Adding up all three figures, we get $p(Y=4)=\frac{29}{384}$.