# Expected values: three fair coins

#### oyth94

##### Member
Suppose you roll three fair coins, and let X be the cube of the number of heads
showing. Compute E(X)

I started this question with X=xi where i = 1,2,3 since there are 3 fair coins and there are either 1 2 or 3 heads that will show up. then it says that X will be the cube of the number of heads so I put
So E(X) = $$\displaystyle xiP(X=xi)$$ = $$\displaystyle i(1/2)^(i^3)$$ summation from i=1 to 3..
am i on the right track??

#### Chris L T521

##### Well-known member
Staff member
Re: expected values: three fair coins

Suppose you roll three fair coins, and let X be the cube of the number of heads
showing. Compute E(X)

I started this question with X=xi where i = 1,2,3 since there are 3 fair coins and there are either 1 2 or 3 heads that will show up. then it says that X will be the cube of the number of heads so I put
So E(X) = $$\displaystyle xiP(X=xi)$$ = $$\displaystyle i(1/2)^(i^3)$$ summation from i=1 to 3..
am i on the right track??
Not quite (and you mean flip 3 fair coins...not roll... XD); since $X$ represents the cube of the number of heads, you instead want $x_1=1^3=1$, $x_2=2^3=8$ and $x_3=3^3=27$. Then $\mathbb{P}(X=x_i)$ would represent the probability of getting 1, 2, or 3 heads; in this case, we would have $\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
$E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots$

I hope this makes sense!

#### oyth94

##### Member
Re: expected values: three fair coins

Not quite (and you mean flip 3 fair coins...not roll... XD); since $X$ represents the cube of the number of heads, you instead want $x_1=1^3=1$, $x_2=2^3=8$ and $x_3=3^3=27$. Then $\mathbb{P}(X=x_i)$ would represent the probability of getting 1, 2, or 3 heads; in this case, we would have $\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
$E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots$

I hope this makes sense!
Oh I see what you did! Thank you so much!

#### mathworker

##### Well-known member
Re: expected values: three fair coins

$\mathbb{P}(X=x_i)=\dfrac{1}{2^i}$, $i=1,2,3$. Thus,
$E[X]=\sum_{i=1}^3 x_i\mathbb{P}(X=x_i)=\ldots$
i have a doubt(probably elementary),
you have assumed probability of getting one head is $$\displaystyle \frac{1}{2}$$.But i think it is the probability to get at least one head ,doesn't that include cases of $$\displaystyle 2$$ or $$\displaystyle 3$$ heads and why isn't the probability $$\displaystyle \frac{1}{2^3}$$ for all cases
(i mean why aren't we taking exact probability here)

#### chisigma

##### Well-known member
Re: expected values: three fair coins

i have a doubt(probably elementary),
you have assumed probability of getting one head is $$\displaystyle \frac{1}{2}$$.But i think it is the probability to get at least one head ,doesn't that include cases of $$\displaystyle 2$$ or $$\displaystyle 3$$ heads and why isn't the probability $$\displaystyle \frac{1}{2^3}$$ for all cases
(i mean why aren't we taking exact probability here)
The probability to have k heads in 3 trials is $\displaystyle \frac{\binom{3}{k}}{8}$, so that...

$P\{k=0\} = \frac{1}{8}$

$P\{k=1\} = \frac{3}{8}$

$P\{k=2\} = \frac{3}{8}$

$P\{k=3\} = \frac{1}{8}$

Kind regards

$\chi$ $\sigma$

#### mathworker

##### Well-known member
got it,changing the positions we could get 3 possibilities of eight , thank you