Let Y be the number of successes in 10 independent Bernoulli trials with probability p of success and X be the number of successes among the first 3 trials.Compute$E[Y|X=x]$
So$X\sim \operatorname{Bin}(3,p) \text{ while } Y\sim \operatorname{Bin}(10,p)$
$\displaystyle E[Y|X=x]\Rightarrow \sum_{y}^{} y \cdot P(Y=y|X=x)=\sum_{y}^{}y \cdot \frac{P(Y=y\cap X=x)}{{P(X=x)}}$
$\displaystyle =\sum_{y=x}^{10}y \cdot\frac{ P(\text{No. of successes in 7 remaining trials}=y-x)}{P(X=x)}$
$\displaystyle=\sum_{y=x}^{10}y \cdot\frac{ {7\choose y-x }{p^{y-x}(1-p)^{7-(y-x)}}}{{3\choose x}{p^x(1-p)^{3-x}}}\Rightarrow \sum_{y=x}^{10}y \cdot\frac{ {7\choose y-x }{p^{y-2x}(1-p)^{4-y+2x}}}{{3\choose x}}$
Am I headed in the right direction? Any idea about how to proceed from here?
What you have done is correct, but it might be tedious to continue the computations.
It is much easier to write $Y = Z_1 + \cdots + Z_{10}$ where each $Z_i$ is a $\text{Bernoulli}(p)$ random variable. Then use linearity of expectation and note that $Z_4, \ldots, Z_{10}$ are independent of $X$, while $E[Z_1 + Z_2 + Z_3 \mid X=x] = x$.