# Probability of tossing a fair coin…

In 5 tosses of a fair coin, let X be the number of heads minus the number of tails. For the probability distribution of X, what is the mean and variance?

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one way to get 5 heads and no tails and one way to get 5 tails and no heads. 5 ways to get 4 heads and 1 tail and 5 ways to get 4 Tails and 1 head. 10 ways to get 3 heads and 2 tails and 10 ways to get 3 tails and one head... does that help give you your distribution? –  Iceman Nov 30 '13 at 4:15

$P(H-T=5)=\frac1{32}$ and $P(H-T=-5)=\frac1{32}$

$P(H-T=3)=\frac5{32}$ and $P(H-T=-3)=\frac5{32}$

$P(H-T=1)=\frac{10}{32}$ and $P(H-T=-1)=\frac{10}{32}$

So the mean $\mu$ would be

$\mu=5\cdot\frac1{32}+3\cdot\frac5{32}+1\cdot\frac{10}{32}-1\cdot\frac{10}{32}-3\cdot\frac5{32}-5\frac1{32}=0$

and the variance $\sigma^2$ would be

$\sigma^2=25\cdot\frac1{32}+9\cdot\frac5{32}+1\cdot\frac{10}{32}+1\cdot\frac{10}{32}+9\cdot\frac5{32}+25\frac1{32}=50\cdot\frac1{32}+18\cdot\frac5{32}+2\cdot\frac{10}{32}=\frac{50+90+20}{32}=\frac{160}{32}=5$

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