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There are two families each having two children. If there are at least two girls among the children, find the probability that all children are girls
(a) $\dfrac{1}{9}$
(b) $\dfrac{1}{10}$
(c ) $\dfrac{1}{11}$
(d) $\dfrac{1}{12}$

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Hint: The probability of a situation is given by $\dfrac{\text{favourable outcome}}{ \text{total outcome}}$. So in order to get the required probability we’ll find how many cases are in our favour and how many total cases we have.

Complete step by step solution: Given, there are two families each having two children and at least two are girls among the children.
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,
$\dfrac{1}{1+4+6}=\dfrac{1}{11}$

Note: In such a type of question we need to case several cases to find the total outcomes. Here at least 2 girls means 2 or more than 2. Counting the total number of cases correctly makes your half job done. Sometimes we take help from topics like permutation and combination to count the cases.