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Q.10.     Find the mean number of heads in three tosses of a fair coin.

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Let X be success of getting head.

When 3  coins are  tossed  then sample space =\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}

\therefore  X can be 0,1,2,3

P(X=0)=P(TTT)=\frac{1}{8}

P(X=1)=P(HTT)+P(TTH)+P(HTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}

P(X=2)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}

P(X=3)=P(HHH)=\frac{1}{8}

The probability distribution is as 

X 0 1 2 3
P(X) \frac{1}{8} \frac{3}{8} \frac{3}{8} \frac{1}{8}

mean number of heads :

                                         =0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}

                                         = \frac{3}{8}+ \frac{6}{8}+ \frac{3}{8}

                                           =\frac{12}{8}

                                            =\frac{3}{2}=1.5

Posted by

seema garhwal

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