Problem 107: Suppose that a cruise ship returns to the United States from the FarEast. Unknown to anyone, 4 of its 600 passengers have contracted a rare disease. Suppose that the Public Health Service screens 20 passengers, selected at random, to see whether the disease is present aboard ship. What is the probability that the presence of the disease will escape detection?
Solution: Let 
be the sample space. The number of outcomes in the sample space is given by
(1) ![\[\binom{600}{20} = \frac{600!}{20! \cdot 580!} . \]](https://1problematatime.com/wp-content/ql-cache/quicklatex.com-3e7476759fdd3c06042f9606b1928735_l3.png "Rendered by QuickLaTeX.com")
Let 
be the event that the a random passenger does not have the disease. That is, the number of outcomes in is given by
(2) ![\[\binom{596}{20} = \frac{596!}{20! \cdot 576!} . \]](https://1problematatime.com/wp-content/ql-cache/quicklatex.com-6771c5dfb84c69a6acc217845669e3e8_l3.png "Rendered by QuickLaTeX.com")
Therefore, the probability of not detecting the disease is
(3) ![\[\frac{\binom{596}{20}}{\binom{600}{20}} \approx 0.87.\]](https://1problematatime.com/wp-content/ql-cache/quicklatex.com-36839a718606015c10ccd1c39f91ae79_l3.png "Rendered by QuickLaTeX.com")
Hence, there is a 87% chance that the disease is not detected.
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