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Question
Find the mean number of heads in three tosses of a fair coin.
Solution
Let X denote the success of getting heads.
Therefore, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
It can be seen that X can take the value of 0, 1, 2 or 3
∴ P(X = 0) = P(TTT)
= P(T) . P(T) . P(T)
= `1/2 xx 1/2 xx 1/2`
= `1/8`
∴ P(X = 1) = P(HHT) + P(HTH) + P(THH)
= `1/2 xx 1/2 xx 1/2 + 1/2 xx 1/2 xx 1/2 + 1/2 xx 1/2 xx 1/2`
= `3/8`
∴ P(X = 2) = P(HHT) + P(HTH) + P(THH)
`1/2 xx 1/2 xx 1/2 + 1/2 xx 1/2 xx 1/2 + 1/2 xx 1/2 xx 1/2`
= `3/8`
∴ P(X = 3) = P(HHH)
= `1/2 xx 1/2 xx 1/2`
= `1/8`
Therefore, the required probability distribution is as follows.
| X | 0 | 1 | 2 | 3 |
| P(X) | `1/8` | `3/8` | `3/8` | `1/8` |
Mean of X E(X), µ =`sum X_iP(X_i)`
= `0 xx1/8 + 1xx3/8 + 2xx3/8 + 3xx1/8`
= `0 + 3/8 + 3/4 + 3/8`
= `12/8`
= 1.5
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