Coin flipping - conditional expectation

Question

We flip a symmetric coin 6 times. Denote X as "amount of all heads gotten" and Y as "count of all heads gotten in the first two flips". Calculate:

• $\mathbb{P}(X=5|Y)$
• $\mathbb{E}(Y|X)$

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I'm learning to my final exam and banging my head against the wall as I already forgot how would one calculate stuff like that. I appreciate any help.

Answer 1

I'll solve for the expectation as it is more fun (for me). I leave the probability as the exercise for the OP :D

Given $X$, we know that we had $X$ heads and $6-X$ tails. Out of symmetry, all arrangements are equally likely, so the probability that the first toss was heads is $\tfrac{X}{6}$. Same goes for the second toss. Since $Y$ is the sum of two indicators (the first and second toss), then

$$E(Y\vert X)=\tfrac{X}{6}+\tfrac{X}{6}=\tfrac{X}{3}$$

Answer 2

Hint

$$\mathbb P\{X=5\mid Y=k\}=\frac{\mathbb P\{X=5, Y=k\}}{\mathbb P\{Y=k\}}.$$ and $$\mathbb E[Y\mid X=k]=\sum_{t\in \{0,1,2\}}t\mathbb P\{Y=t\mid X=k\}.$$