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Expected values: three fair coins

oyth94

Member
Jun 2, 2013
33
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
P(getting heads) = (1/2)^(i^3)
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
Jan 26, 2012
995
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
P(getting heads) = (1/2)^(i^3)
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
Jun 2, 2013
33
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

Active member
May 31, 2013
118
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
Feb 13, 2012
1,704
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

Active member
May 31, 2013
118
got it,changing the positions we could get 3 possibilities of eight , thank you