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I'm trying to understand the solution to a problem from A First Course in Probability (Ross):

There are two misshapen coins in a box; their probabilities for landing on heads when they are flipped are, respectively, .4 and .7. One of the coins is to be randomly chosen and flipped 10 times. Given that two of the first three flips landed on heads, what is the conditional expected number of heads in the 10 flips?

Here is the solution provided in the book: enter image description here

Intuitively, the solution makes sense to me. We want to use the information that two of the first three flips were heads to update our beliefs of which coin was chosen. However, I don't understand what's happening under the hood. IIUC, $N_7, C$ are random variables, but is $T$ a RV as well given we know that it occurred. It seems like the underlying formula used by the solution is:

(1) $E[N_7|T] = E[E[N_7|T, C]]$, where the outer expectation is over $C$, is that correct?

If so, I would have expected (1) to decompose into:

$E[N_7|T] = \sum_c E[N_7|T, C=c]p(c)$ but this seems different that what the solution is proposing.

I think I'm overlooking/missing some key properties of expectations/RVs/bayes theorem here. Any insight would be appreciated, thanks!

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    $\begingroup$ $T$ is not a random variable (it doesn't take on numerical values as random variables do); it is an event which either occurs or does not occur. Also, there is a typo in the solution: in the formula for $E[N_7\mid T]$, the $P(C_1T]$ should be $P(C_1\mid T)$ $\endgroup$ Commented Jul 25, 2022 at 13:47
  • $\begingroup$ That makes sense, yes I saw the typo as well. $\endgroup$
    – Marko
    Commented Jul 25, 2022 at 14:21

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It suspect you have gotten confused in the "decomposition" of your equation (1).

In order to try to make things clear and bring out the underlying structure, I'm supplying a somewhat abstract "explanation" and hoping you can apply it to your case with your notation. I apologize if this is not helpful.

First, \begin{equation} E[x|z] = \sum_x x\,p(x|z) . \end{equation} Second, \begin{equation} p(x|z) = \sum_y p(x,y|z) = \sum_y p(x|y,z)\,p(y|z) . \end{equation} Therefore, \begin{equation} E[x|z] = \sum_x x\sum_y p(x|y,z)\,p(y|z) . \end{equation}

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