Let's say there are 5 different diseases; 1/100 people have one of these five diseases.
Just to be clear, what I means is: 1/100 people have the first disease, 1/100 people have the second disease, 1/100 people have the third disease, 1/100 people have the fourth disease, and 1/100 people have the fifth disease.
If you have a group of 500 randomly selected individuals, what is the probability that at least one person has one of the five diseases?
The way I understand it, since there are 5 diseases each with 1/100 probability, then 5/100 people have one of these diseases.. So, if you have 500 people, the probability that someone has the disease would be 500*(5/100) = 25. But this makes no sense! What am I doing wrong here? Thank you.
EDIT: A person cannot have more than one of the five diseases.
Best Answer
You are missing a few important assumptions in the statement of the problem. If it turns out that the probability of getting each disease individually are dependent and not mutually exclusive, then there is not enough information to solve.
If you are not able to be sick with two diseases simultaneously (the diseases are mutually exclusive), then 5 in 100 people will be sick on average (with exactly one disease). Out of 500 people, the probability that at least one person is sick is opposite the probability that none of them are sick.
The probability none are sick is: $\binom{500}{0}(\frac{95}{100})^{500}(\frac{5}{100})^0 = (\frac{95}{100})^{500}$, so the probability at least one is sick is $1 - (\frac{95}{100})^{500}$
The other possible interpretation of the problem is that it is possible to get multiple diseases at once, and the chances of doing so are independent of one another. The chance someone is healthy then is $(\frac{99}{100})^5$.
So, the probability none are sick is $\binom{500}{0}((\frac{99}{100})^{5})^{500}((\frac{1}{100})^{5})^{0} = ((\frac{99}{100})^{5})^{500}$
Thus, the probability that at least one is sick is $1-((\frac{99}{100})^{5})^{500}$