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(a) $\dfrac{1}{9}$

(b) $\dfrac{1}{10}$

(c ) $\dfrac{1}{11}$

(d) $\dfrac{1}{12}$

Answer
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The total number of children in the families is 4.

Let us take the following cases:

Case I:

There are 2 girls and 2 boys in the families.

As we have to choose 2 boys thus we can choose in $^{4}C_{2}$ways.

Case II:

There are 3 girls and 4 boys in the family.

As we have to choose 1 boy thus we can choose in $^{4}C_{1}$ways.

Case III:

There are 4 girls and 0 boys in the families.

Thus we can choose in only 1 way.

The favourable outcome is 1.

The total outcome is ${1}+{}^{4}{C_{1}}+{}^{4}{C_{2}}$

The probability that all children are girls is

$\dfrac{1}{{1}+{}^{4}{C_{1}}+{}^{4}{C_{2}}}$

By simplifying using the formula ${}^n{C_r}=\dfrac{n!}{r! \times \left(n-r!\right)}$ the combination we get,

$\dfrac{1}{1+4+6}=\dfrac{1}{11}$