If the probability of a male child is $\frac{1}{2}$ , the probability that in a family of 4 children there will be atleast one boy and one girl is

\frac{3}{8}

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\frac{5}{8}

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\frac{7}{8}

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\frac{1}{8}

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Solution

The correct option is D $\frac{7}{8}$
$P$ (at least $1$ boy, $1$ girl) $= 1 - P (n o G i r l) - P (n o B o y)$
$=> - \frac{1}{2^{4}} - \frac{1}{2^{4}} = \frac{7}{8}$

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Assume that each child born is equaly likely to be boy or a girl. IF a family has two children, what is the conditional probability that both are girls given that
atleast one is a girl?

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If the probability of a male child is 12, the probability that in a family of 4 children there will be atleast one boy and one girl is

The correct option is D 78P(at least 1 boy, 1 girl) =1−P(no Girl)−P(no Boy)=>−124−124=78

If the probability of a male child is $\frac{1}{2}$ , the probability that in a family of 4 children there will be atleast one boy and one girl is

The correct option is D $\frac{7}{8}$
$P$ (at least $1$ boy, $1$ girl) $= 1 - P (n o G i r l) - P (n o B o y)$
$=> - \frac{1}{2^{4}} - \frac{1}{2^{4}} = \frac{7}{8}$