You need the hypergeometric distribution which will give the needed probabilities.
There are only 2 choices: 1 woman or 2 women.
The probability of 1 woman is 15/28 and of 2 women is 3/28.
The expectation is (15/28) + (6/28) = 21/28= 3/4
India G.
asked 12/07/18From a group of 3 women and 5 men, a delegation of 2 is selected. Find the expected
number of women in the delegation
You need the hypergeometric distribution which will give the needed probabilities.
There are only 2 choices: 1 woman or 2 women.
The probability of 1 woman is 15/28 and of 2 women is 3/28.
The expectation is (15/28) + (6/28) = 21/28= 3/4
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