Question

Three groups of children contain $3$ girls and $1$ boy; $2$ girls and $2$ boys; $1$ girl and $3$ boys. One child is selected at random from each group. Find the chance that the three selected children consist of $1$ girl and $2$ boys.
(A) $\dfrac{{13}}{{36}}$ (B) $\dfrac{{13}}{{32}}$ (C) $\dfrac{{15}}{{32}}$ (D) $\dfrac{{15}}{{36}}$

Answer 1

Hint: The selected child from each group can be either a boy or a girl. Form different cases and solve each case separately then use the rule of probability of mutually exclusive events.

Complete step-by-step answer:
Total number of children in the first group is $4$, in the second group is $4$ and in the third group is also $4$.
First group consists of $3$ girls and $1$ boy.
Second group consists of $2$ girls and $2$ boys.
Third group consists of $1$ girls and $3$ boys.
We have to select $1$ child from each group in such a way that the selected children consist of $1$ girl and $2$ boys. Based on the given information, three cases are possible:
Case $1$: We have selected $1$ boy from the first group, $1$ boy from the second group and $1$ girl from the third group. The required probability in this case will be:
$
   \Rightarrow {P_1} = \dfrac{1}{4} \times \dfrac{2}{4} \times \dfrac{1}{4}, \\
   \Rightarrow {P_1} = \dfrac{1}{{32}} \\
$
Case $2$: We have selected $1$ boy from first group, $1$ girl from second group and $1$ boy from third group. The required probability in this case will be:
$
   \Rightarrow {P_2} = \dfrac{1}{4} \times \dfrac{2}{4} \times \dfrac{3}{4}, \\
   \Rightarrow {P_2} = \dfrac{3}{{32}} \\
$
Case$3$: We have selected $1$ girl from first group, $1$ boy from second group and $1$ boy from third group. The required probability in this case will be:
$
   \Rightarrow {P_3} = \dfrac{3}{4} \times \dfrac{2}{4} \times \dfrac{3}{4}, \\
   \Rightarrow {P_3} = \dfrac{9}{{32}} \\
$
And all the above cases are mutually exclusive. So, the total probability will be the sum of all the probabilities of the three cases:
$
   \Rightarrow P = {P_1} + {P_2} + {P_3}, \\
   \Rightarrow P = \dfrac{1}{{32}} + \dfrac{3}{{32}} + \dfrac{9}{{32}}, \\
   \Rightarrow P = \dfrac{{13}}{{32}}. \\
$
Thus, the chance of the three selected children consisting of $1$ girl and $2$ boys is $\dfrac{{13}}{{32}}$. (B) is the correct option.

Note: If $A$ and $B$ are two mutually exclusive events with the probability of their occurrence being $P\left( A \right)$ and $P\left( B \right)$ respectively, then the probability of occurrence of either of them is:
$ \Rightarrow P\left( {A{\text{ or }}B} \right) = P\left( A \right) + P\left( B \right)$.