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Find the mean number of heads in three tosses of a fair coin.

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#### 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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