Cfrgtkky

Suppose you roll one fair six-sided die and then flip as many coins as the number showing on the die. (For example, if the die shows 4, then you flip four coins.) Let Y be the number of heads obtained. Compute $E(Y)$.

So, $E(Y)=Sigma_{y=1}^{y=6}yP_{y}(y)$

We will define $X$ as the number on the die. Therefore:

$$E(Y)=sum_{y=1}^{y=6}yP_{y}(y)=sum_{y=0}^{y=6}sum_{x=1}^{x=6}(y)P(X=x,Y=y)=sum_{y=0}^{y=6}ysum_{x=1}^{x=6}(1/6)binom{x}{y}(1/2)^x$$

This inner equation is valid: $(1/6)binom{x}{y}(1/2)^x$

Basically, it says that if I roll a 3, and I get 1 head only, what is the probability of that occurring = $(1/6)text{[for the roll]}binom{3}{1}text{[from 3 rolls, we choose one head]}(1/2)^3=1/16$

Now, I need to simplify the above double sum, but I am not sure how to do it. The answer in my textbook is $7/4$, but it does not show steps. I think I am right until now, I just need help doing this summation.

If $D$ is the result of the die-throw, $E(Y) = sum_{d=1}^6 E(Y|D=d)P(D=d)$.

Clearly $E(Y|D=d)= frac{d}{2}$. So we are left with $$E(Y) = sum_{d=1}^6 frac{1}{6}frac{d}{2} = frac{1}{12}sum_{d=1}^6 d = frac{21}{12}$$