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Suppose you flip four coins and you're guaranteed that at least 2 are heads. What is the expected value of the number of heads? I have the following calculation.

\begin{align} & \operatorname E\left[ X \mid X \geq 2\right] \\[8pt] = {} & \sum_{n = 2}^4 nP [X=n\mid X \geq 2] \\[8pt] = {} & 2 \cdot \frac{\binom{4}{2}}{11} + 3 \cdot \frac{4}{11} + 4 \cdot \frac{1}{11} \\[8pt] = {} & 2.5454. \end{align}

Is this calculation correct? Specifically is the use of the conditional probability distribution in the first step correct?

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    $\begingroup$ Looks good to me! $\endgroup$ – Matthew Daly Dec 8 '19 at 21:06
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\begin{align} \mathbb E[X\mid X\geqslant 2] &= \frac{\mathbb E[X\mathsf 1_{\{X\geqslant 2\}}]}{\mathbb P(X\geqslant 2)}\\ &= \frac{\sum_{k=2}^4 k\binom 4k 2^{-4}}{\sum_{k=2}^4\binom 4k 2^{-4}}\\ &= \frac{\sum_{k=2}^4 k\binom 4k}{\sum_{k=2}^4\binom 4k}\\ &= \frac{28}{11}. \end{align}

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The argument given in the question is correct. (If I were writing it to be read by someone who doesn't know how to do this until they read what I wrote, I would probably say more than what you've said.)

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