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Find the mean and variance of the number of heads when two coins are tossed simultaneously.

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Correct Answer - Mean =1, variance = 0.5 s={HH, HT, TH, TT}. Let X be the number of heads. Then, X=0, 1 or 2. P(X=0)=P(getting no head) `=(1)/(4).` P(X=1)=P(getting 1 head)`=(2)/(4)=(1)/(2).` ltbtgt P(X=2)=P(getting 2 heads) `=(1)/(4).`

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