Question

Three groups of children contain respectively 3 girls & 1 boy; 2 girls & 2 boys; 1 girl & 3 boys. One child is selected at random from each group. What is the chance that the three selected children consist of one girl and 2 boys?
A. $\dfrac{9}{{32}}$
B. $\dfrac{{11}}{{32}}$
C. $\dfrac{{13}}{{32}}$
D. $\dfrac{7}{{32}}$

Answer 1

Hint: We just focus on what we have to calculate here, we have to select three children of which one is a girl and two are boys. We have only three groups here. For that we can see that we have only three possibilities here i.e. we can have a girl from group first, a boy from second group and a boy from third group or we can have a boy from group first, a girl from second group and a boy from third group or we can have a boy from group first, a boy from second group and a girl from third group. We will calculate all the probabilities and just have a union of all of them because we have the or condition here between the possibilities and the events are mutually exclusive i.e., they cannot occur at the same time.

Complete step-by-step answer:
Let us suppose group of 3 girls & 1 boy be group A, group of 2 girls & 2 boys be group B and group of 1 girl & 3 boys be group C.
For the combination we have only three possibilities here to select the combination of one girl and 2 boys at random.
We know that, Probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes.
First Possibility,
A girl from group A, a boy from group B, a boy from group C
To select a girl from Group A, we can select 3 out of 4 possibilities, so probability $= \dfrac{3}{4}$
To select a boy from Group B, we can select 2 out of 4 possibilities, so probability $= \dfrac{2}{4}$
To select a boy from Group C, we can select 3 out of 4 possibilities, so probability $= \dfrac{3}{4}$
So, the probability to select a girl from group A, a boy from group B, a boy from group C $= \dfrac{3}{4} \times \dfrac{2}{4} \times \dfrac{3}{4} = \dfrac{9}{{32}}$
Second Possibility,
A boy from group A, a girl from group B, a boy from group C
To select a boy from Group A, we can select 1 out of 4 possibilities, so probability $= \dfrac{1}{4}$
To select a girl from Group B, we can select 2 out of 4 possibilities, so probability $= \dfrac{2}{4}$
To select a boy from Group C, we can select 3 out of 4 possibilities, so probability $= \dfrac{3}{4}$
So, the probability to select a boy from group A, a girl from group B, a boy from group C $= \dfrac{1}{4} \times \dfrac{2}{4} \times \dfrac{3}{4} = \dfrac{3}{{32}}$
Third Possibility,
A boy from group A, a boy from group B, a girl from group C
To select a boy from Group A, we can select 1 out of 4 possibilities, so probability $= \dfrac{1}{4}$
To select a boy from Group B, we can select 2 out of 4 possibilities, so probability $= \dfrac{2}{4}$
To select a girl from Group C, we can select 1 out of 4 possibilities, so probability $= \dfrac{1}{4}$
So, the probability to select a boy from group A, a boy from group B, a girl from group C $= \dfrac{1}{4} \times \dfrac{2}{4} \times \dfrac{1}{4} = \dfrac{1}{{32}}$
Now, we know that these three possibilities are mutually exclusive, so our required probability is the sum total of all the probabilities.
Required probability is the sum of probability of first possibility, probability of second possibility and probability of third possibility.
Required probability $= \dfrac{9}{{32}} + \dfrac{3}{{32}} + \dfrac{1}{{32}} = \dfrac{{13}}{{32}}$
Hence our required probability is $\dfrac{{13}}{{32}}$
Option C. $\dfrac{{13}}{{32}}$ is our correct answer.

Note: In such a type of problem just take into consideration that probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. We have to check for all the possibilities of the events and then we have to check if the events are mutually exclusive.
When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event i.e. $P(AorB) = P(A) + P(B)$
In this question they are mutually exclusive and for the mutually exclusive events they cannot happen at the same time and we just take the sum total of the individual probabilities to get the total probability of the event.