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Discrete Random Variables
we are working with commonly used discrete and continuous distribution In a hospital there are
16 patients, 4 of whom have AIDS. A doctor is assigned to 6 of these patients at random. What is
the probability that he gets 2 of the AIDS patients?
Solution
There are 16C6 ways to choose any 6 patients.
There are 4C2 ways to get 2 with AIDS, and 12C4 to choose 4 without.
Thus,
P = [(12C4)(4C2)]/(16C6) = 0.370879121 [ANSWER]

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  • 1. Discrete Random Variables we are working with commonly used discrete and continuous distribution In a hospital there are 16 patients, 4 of whom have AIDS. A doctor is assigned to 6 of these patients at random. What is the probability that he gets 2 of the AIDS patients? Solution There are 16C6 ways to choose any 6 patients. There are 4C2 ways to get 2 with AIDS, and 12C4 to choose 4 without. Thus, P = [(12C4)(4C2)]/(16C6) = 0.370879121 [ANSWER]