a group of people?

Question

from a group of 20 men and 16 women 2 people are chosen at random. find the probability of
a) 1 man 1 woman being chosen
b) 2 men being chosen
c) 2 women being chosen

Answer 1

Since there are 20 men and 16 women, the total number of people is 20+16=36, so the probability of choosing a man is 20/36=10/18=5/9, and the probability of choosing a woman is 16/36=8/18=4/9.

a) Since we are looking for exactly one man and one woman, the first person selected could be of either gender. We can meet the criteria if our selection is either MW or WM. So the probability of one man and one woman is P(MW)+P(WM). The probability of man-woman is (5/9)x(4/9)=20/81, and the probability of woman-man is (4/9)x(5/9)=20/81. So the probability of one man and one woman is (20/81)+(20/81)=40/81.

b) The probability of choosing two men is (5/9)x(5/9)=25/81.

c) The probability of choosing two women is (4/9)x(4/9)=16/81.

Since these are the only three possibilities for selecting two people at random, the sum of them should add up to 1. And indeed, (40/81)+(25/81)+(16/81)=1.

Answer 2

Do a binomial expansion with p=5/9 (man) and q= 4/9=(woman) of (p+q)^2. The first term is two men chosen, the middle term is one of each, the last term is two women chosen.

Answer 3

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