Question

1. A class has 10 boys and 5 girls. Three students are selected at random one after another. Find the probability that
(i) first two are boys and third is girl (ii) first and third are of same sex and the second is of opposite sex.

Answer 1

Answer on Question #53389 – Math – Statistics and Probability

A class has 10 boys and 5 girls. Three students are selected at random one after another. Find the probability that

(i) the first two are boys and the third is girl;

(ii) first and third are of the same sex and the second is of opposite sex.

Solution

If a class has 10 boys and 5 girls, and three students are selected at random one after another then

(i) the probability, that the first two are boys and third is girl, is

$$p = \frac{10}{15} \cdot \frac{9}{14} \cdot \frac{5}{13} = \frac{30}{182} = \frac{15}{91} \approx 0.165;$$

(ii) the probability, that the first and the third are of the same sex and the second is of opposite sex, is

$$\begin{array}{l} p = p(1 = \text{boy}, 2 = \text{girl}, 3 = \text{boy}) + p(1 = \text{girl}, 2 = \text{boy}, 3 = \text{girl}) = \\ = \frac{10}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} + \frac{5}{15} \cdot \frac{10}{14} \cdot \frac{4}{13} = \\ = \frac{10 \cdot 9 \cdot 5 + 10 \cdot 5 \cdot 4}{15 \cdot 14 \cdot 13} = \frac{10 \cdot 9 + 10 \cdot 4}{3 \cdot 14 \cdot 13} = \frac{5 \cdot 9 + 5 \cdot 4}{3 \cdot 7 \cdot 13} = \frac{65}{3 \cdot 7 \cdot 13} = \frac{5}{3 \cdot 7} = \frac{5}{21} \approx 0.238. \end{array}$$

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