My friends and I posed this question and I'm really curious how something like this would be calculated. I'm not sure if I phrased it correctly in the title, but increasingly complex examples I came up with would be:
If person 1 has a 1.5% chance of catching a cold each time they interact with person 2, who has a 5% chance of already having a cold, what are the odds of person 1 catching a cold each time they interact with person 2?
If two people have a 1.5% chance of catching a cold each time they interact with others, and person 1 interacted with 50 people before interacting with person 2, what are the odds that person 2 could catch a cold each time they interact with person 1?
If three people people all have a 1.5% chance of catching a cold each time they interact with others, person 1 interacted with 50 people, then person 2 interacted with person 1 10 times, what are the odds of person 3 catching a cold each time they interact with person 2?
If person 1 has a 5% chance of already having a cold and persons 2 and 3 both have a 1.5% chance of catching a cold each time they interact with others, what are the odds of person 3 catching a cold each time they interact with person 2 if person 2 previously interacted with person 1 10 times?
Your first question...
If person 1 has a 1.5% chance of catching a cold each time they interact with person 2, who has a 5% chance of already having a cold, what are the odds of person 1 catching a cold each time they interact with person 2?
P(catch cold | $n$ meetings) = P(you catch cold on $n$ meetings given he has cold) x P(he has cold)
$$ P(C|n) = (.05)(1-(1-.015)^n) $$
Next
If two people have a 1.5% chance of catching a cold each time they interact with others, and person 1 interacted with 50 people before interacting with person 2, what are the odds that person 2 could catch a cold each time they interact with person 1?
Let $C_1$ denote the event the first guy has a cold.
$$ P(C_1)=1-.985^{50} $$ Then the event second guy has cold, $C_2$, after $m$ interactions has the probability $$ P(C_2)=1-(1-P(C_1)^{m}) $$ or $$ P(C_2)=1-(1-(1-.985^{50}))^{m} $$