Question

Three groups of children contain 3 girls and one boy; 2 girl and 2 boys, one girl and 3 boys. One child is selected at random from each group. Show that the chance that the three selected consist of 1 girl and 2 boys is

• A. $\frac{13}{36}$
• B. $\frac{13}{32}$
• C. $\frac{15}{32}$
• D. $\frac{15}{36}$

Answer 1

The correct option is B $\frac{13}{32}$

We tabulated the given data as follows:

	Boys	Girls
Group 1	3	1
Group 2	2	2
Group 3	1	3

out of each group , 1 child is related. The three selected can consist of 1 girl and 2 boys in the follows three mutually exclusive ways:
i) 1 boy from Group 1, 1 boy from Group 2 , 1 girl from group 3.
ii) 1 boy from group 1,1 girl from group 2, 1 boy from group 3.
iii) 1 girl from group 1, 1 boy from group 2, 1 boy from group 3.
In calculating probabilities below it is noted that choosing a child from any of the three group is in dependent of choosing a child in the other groups.
Hence
$P\left(i\right)=\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}=\frac{9}{32}$
$P\left(ii\right)=\frac{3}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{3}{32}$
$P\left(iii\right)=\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{1}{32}$
So, the required probability is equal to
$P\left(i\right)+P\left(ii\right)+\left(iii\right)=\frac{9}{32}+\frac{3}{32}+\frac{1}{32}=\frac{13}{32}$