Assuming that for a husband-wife couple,the c | vedclass

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(C) The probability of having a boy is $p = \frac{1}{2}$ and the probability of having a girl is $q = \frac{1}{2}$.For $n = 2$ children,we want to fin

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$\frac{1}{2}$

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(C) The probability of having a boy is $p = \frac{1}{2}$ and the probability of having a girl is $q = \frac{1}{2}$.For $n = 2$ children,we want to find the probability of having exactly $1$ boy and $1$ girl.Using the binomial distribution formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$,where $n = 2$ and $k = 1$:$P(1 	ext{ boy, } 1 	ext{ girl}) = {}^2C_1 \cdot \left(\frac{1}{2}ight)^1 \cdot \left(\frac{1}{2}ight)^{2-1}$$P(1 	ext{ boy, } 1 	ext{ girl}) = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{2}{4} = \frac{1}{2}.$

Assuming that for a husband-wife couple,the chances of their child being a boy or a girl are the same,the probability of their two children being a boy and a girl is:

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