Question

A class contains 10 men 20 women of which half of men and half of the women have brown eyes a person is selected one by one find the probability that
1. First one has brown eye and second and third are not
brown eyes
2. First two are women and third and forth are men

Answer 1

1. There are a total of 15 brown-eyed people in the class, so the probability that the first chosen one is brown-eyed is ${p_1} = \frac{{15}}{{30}} = \frac{1}{2}$ .

The probabilities that the second and third chosen are not brown-eyed are equal, respectively

${p_2} = \frac{{15}}{{29}},\,\,{p_3} = \frac{{14}}{{28}} = \frac{1}{2}$ .

So, the wanted probability is $p = {p_1}{p_2}{p_3} = \frac{1}{2} \cdot \frac{{15}}{{29}} \cdot \frac{1}{2} = \frac{{15}}{{116}}$

Answer: $p = \frac{{15}}{{116}}$

2.. The probabilities that the first and second chosen are women are equal respectively

${p_1} = \frac{{20}}{{30}} = \frac{2}{3},\,\,{p_2} = \frac{{19}}{{29}}$

The probabilities that the third and fourth chosen are men are equal respectively

${p_3} = \frac{{10}}{{28}}=\frac{{5}}{{14}},\,\,{p_4} = \frac{{9}}{{27}}= \frac{{1}}{{3}}$

So, the wanted probability is $p = {p_1}{p_2}{p_3}{p_4} = \frac{2}{3} \cdot \frac{{19}}{{29}} \cdot \frac{{5}}{{14}} \cdot \frac{{1}}{{3}} = \frac{{95}}{{{\rm{1827}}}}$

Answer: $p = \frac{{95}}{{{\rm{1827}}}}$