Question

Imagine that 60 passengers on a cruise ship might be carrying a rare virus, and government health workers randomly (without replacement) select ten passengers for testing. What is the minimum number of the 60 passengers that must be carrying the virus to make the probability that at least one of the ten selected passengers tests positive for the virus is greater than or equal to 0.95?

Answer 1

15

Explanation:

Lets say that P passengers are infected with the virus.

We want the probability of at least one of the ten passengers to be infected, so first we need to calculate the probability of no one of the 10 passengers to be infected, and make one minus this probability (the opposite of none being infected is that at least one person is infected)

picking the first person, the probability of this person not being infected is (60-P)/60

then, picking the second person, the probability is (59-P)/59 and so on

doing this for 10 passengers, the probability of at least one passengers of the 10 being infected is 1 - [(60-P)/60]*[(59-P)/59]*...*[(51-P)/51]

This probability needs to be greater than or equal to 0.95, so:

1 - [(60-P)/60]*[(59-P)/59]*...*[(51-P)/51] >= 0.95

[(60-P)/60]*[(59-P)/59]*...*[(51-P)/51] <= 0.05

For P = 1, we have

59*58*...*50/(60*59*...*51) = 50/60 = 0.8333

For P=2, we multiply the probability found with P=1 by 49/59, then for P=3, multiply the previous result by 48/58, and so on until we find a value lesser or equal than 0.05

For P = 10, we have

50*49*...*41/(60*59*...*51) = 0.1362

For P = 14, we have

46*44*...*37/(60*59*...*51) = 0.0541

For P = 15, we have

45*44*...*36/(60*59*...*51) = 0.0423

So, the value of P we want is 15