Question 1078048: Suppose that a cruise ship return to the Philippines from the Bahamas. Unknown to anyone, four of its 600 passengers have contracted a rare disease. Suppose that DOH screen 20 passengers, selected at random, to see whether the disease is present aboard the ship. What is the probability that the presence of the disease will escape detection?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The number of possible samples is 600C20.
None would be 596C20
For 1 it would be 4C1*596C19=0.1210; 4 ways to choose the affected passenger, 596C19 the number of ways to select 19 from 596.
For 2 it would be 4C2*596C18=0.006
For 3 it would be 4C3*596C17=0.0001
0.1219 would see it and 0.8781 would not see it. Note, one could multiply (596/600)*(595/599)*...*(577/581) as well, which gives 0.8729, and likely with some error, since by fractions, there has to be rounding and by multiplying all the numerator and denominator terms, one is dealing with 10^56.
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