a symmetrical die is thrown, then a coin is thrown as many times as the die indicates .

Question

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Exercices :

a symmetrical die is thrown, then a coin is thrown as many times as the die indicates points.

let $X$ be the number of Tails obtained .

Determine $E(X)$

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My attempts :

First it's clearly that : $$X(\Omega)=\left(1;2;3;4;5;6 \right)$$

Now I need calculs $P(X=1)$ so number of die is $1$ So we thrown the coin $1$-time So : $$P(X=1)=\frac{1}{6}×\frac{1}{2}=\frac{1}{12}$$ Also : $$P(X=2)=\frac{1}{6}×\frac{1}{2}+\frac{1}{6}×\frac{1}{2}=\frac{1}{6}$$ Is my work correct !!

I need to se more ideas and excellent way

• Edit :

Any solution( answer)!!

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Thanks

Answer 1

You are somehow on the right track, but your equations are wrong. Let $D$ be the result of the die throw. Then you have actually computed $$ \newcommand{\E}{\Bbb E}\newcommand{\P}{\Bbb P} E(X\mid D=1) P(D=1) =\frac{1}{12},$$$$E(X\mid D=2)P(D=2)=\frac16, $$ and so on. If you sum all these, you get the answer by the law of total expectation: $$ E(X)=\sum_{i=1}^6E(X\mid D=i)P(D=i). $$ We can also get away with even less calculation, because the above is equal to $$ \sum_i \frac i2P(D=i) = \frac{E(D)}{2}. $$ I hope you’ll do your best to understand everything that’s going on here, because it’s quite instructive for similar problems.

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By the way, if you want to directly compute $P(X=1)$ for example, it’s much more messy: $$ P(X=1)=\sum_iP(X=1\mid D=i)P(D=i) $$$$ =\frac12\cdot\frac16 + \frac24\cdot\frac16+\frac38\cdot\frac16+\ldots . $$ And it only gets worse with higher values than $1$ - you’d have to use binomial coefficients all over the place.