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(B) The probability of a boy being born is $P(B) = \frac{1}{2}$ and the probability of a girl being born is $P(G) = \frac{1}{2}$.In a family of $4$ ch

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$\frac{15}{16}$

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(B) The probability of a boy being born is $P(B) = \frac{1}{2}$ and the probability of a girl being born is $P(G) = \frac{1}{2}$.In a family of $4$ children,the total number of outcomes is $2^4 = 16$.The event of having 'at least one girl' is the complement of the event of having 'no girls' (i.e.,all boys).The probability of having no girls is $P(	ext{all boys}) = (\frac{1}{2})^4 = \frac{1}{16}$.Therefore,the probability of having at least one girl is $1 - P(	ext{all boys}) = 1 - \frac{1}{16} = \frac{15}{16}$.

If the probabilities of a boy and a girl being born are the same,then in a $4$ children family,the probability of having at least one girl is:

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