Conditional expectation value

Question

A taxi company has $n$ cars each with two mirrors. Each mirror breaks off independently with probability $p$, let $D$ denote the number of cars that have lost both mirrors and $M$ the total number of broken mirrors.
Find: $$ E(D|M=m) \quad \text{where $m=0,...,2n$} $$ So far I have tried to express:
$$ P(D=k|M=m)=\frac{P(D=k;M=m)}{P(M=m)} =\frac{P(M=m|D=k)P(D=k)}{P(M=m)} $$ Now here I'm stuck because I tried to express $P(M=m|D=k)$ as a sum but but I get very difficult expressions to simplify. And I can't really compute the sum for the expected value.

Answer 1

Find $\mathsf E(D\mid M=m)$

Hint: Use Indicator Random Variables and the Linearity of Expectation.

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Spoilers: Don't unwrap until Christmas. No, peeking until you've tried it yourself.

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Let $D_i$ be the event that car $i$ has both mirrors broken, and $\mathbf 1_{D_i}$ the indicator random variable of that event.   (Meaning it has value $1$ when $D_i$ and $0$ otherwise.)

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Then by the linearity of expectation and the identical (conditional) distribution of the events : $$\begin{align}\mathsf E(D\mid M=m) & = \mathsf E(\sum_{i=1}^n\mathbf 1_{D_i}\mid M=m) \\ &= \sum_{i=1}^n\mathsf P(D_i\mid M=m)\\ &= n~\mathsf P(D_1\mid M=m)\end{align}$$

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So, what is the conditional probability that a particular car has both mirrors broken when given that $M$ of the $2n$ mirrors are broken?