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.The chance that the three selected children consists of one girl and 2 boys is

• A. $\frac{9}{32}$
• B. $\frac{11}{32}$
• C. $\frac{13}{32}$
• D. $\frac{7}{32}$

Answer 1

Answer: (C)

$\frac{13}{32}$

Let:
Group A: 3 girls, 1 boy
Group B: 2 girls, 2 boys
Group C: 1 girl, 3 boys
The combination of 1 girl and 2 boys can be selected in:
(i) A - girl, B - boy, C - boy: $\frac{3}{4}\times \frac{2}{4}\times \frac{3}{4}=\frac{18}{64}$
(ii) A - boy, B - girl, C - boy: $\frac{1}{4}\times \frac{2}{4}\times \frac{3}{4}=\frac{6}{64}$
(iii) A - boy, B - boy, C - girl: $\frac{1}{4}\times \frac{2}{4}\times \frac{1}{4}=\frac{2}{64}$
Hence, required probability = $\frac{18}{64}+\frac{6}{64}+\frac{2}{64}=\frac{26}{64}=\frac{13}{32}$