A 10-yen coin is thrown $N$ times, where $N$ is the number of total heads of obtained by tossing three $100$-yen coins together. Calculate the expected value of the number of total heads obtained by tossing the $10$ yen coin.
My attempt Let $10$ yen coin be thrown $N$ times $= X_1+X_2+X_n$
$E(\sum_{i=1}^{N} X_i)\mbox{ for } i=1,2,3 $
$P(N=k)$ is for total of head when three 100 yen coin is tossed.
$$\begin{aligned} P(N=0) &= \frac18 \\ P(N=1) &= \frac38 \\ P(N=2) &= \frac38 \\ P(N=3) &=\frac18 \\ P(N=k) &\approx \frac32 \end{aligned} \\ P(\text{number of total 100 yen head})\cdot E(\sum_{i=1}^{N} X_i\mid P(N=k)) \\ E(\sum_{i=1}^{N} X_i)=P(N=k) \cdot E(\sum_{i=1}^{N} X_i\mid P(N=k)$$
Is this right? What should I do next? I'm quite confused, of the relation of N and expectation
Answer is $3/4$, but how can I find it?
I computed expectation, $N$ for $10$ yen is $$\begin{aligned} N &= 0~, & 0.5 \cdot 0 &= 0 \\ N &= 1~, & 0.5 \\ N &= 2~, & 0.5 \cdot 2 & \\ N &= 3~, & 0.5 \cdot 3 & \end{aligned} \\ P(\text{number of total 100 yen head}) \cdot E(\sum_{i=1}^{N} X_i\mid P(N=k)) = 0 \cdot \frac18 + \frac3 8 \cdot 0.5 + \frac38 \cdot1 + \frac18 \cdot 1.5$$ am I right??
Let $X$ be the count of heads on the $N$ coins, where $N$ is the count of heads on tossing three coins (all coins being fair and independent). So $N\sim\mathcal{Bin}(3,1/2)$ and $X\mid N\sim\mathcal{Bin}(N,1/2)$.
We seek $\mathsf E(X)$ and by the Law of Total Probability: $$\mathsf E(X)=\mathsf E(\mathsf E(X\mid N))$$
The expectation for a Binomially distributed random variable is well known, so this will be easy enough for you to evaluate.