I cannot wrap my head around this scenario of having 'at least 2 of 3', I've been stuck with this question and I can only understand that for 'at least 1 of 2' you can approach it with complement of it never occurring…
At first I thought I could do "$P(A)+P(B)+P(C)-P(A\cap B\cap C)$", but that doesn't make much sense.
How should I approach this question? what method do I use?
A national polling company polled Canadians about their coffee consumption and made the following discovery: 81% of Canadians have at least one cup of coffee a day. Two Canadians are randomly chosen on a randomly chosen day. What is the probability
Suppose you are to randomly inspect three Canadians on a randomly chosen day. Find the probability that at least two of the three has consumed at least one cup of coffee.
Best Answer
You have a binomial distribution with the individual probability being $0.81$. You can either compute the chance that exactly two drank coffee and add to the chance all three drank coffee, or you can compute the chances that none and exactly one drank coffee and subtract from $1$. In this case the amount of computation is almost the same. The chance that exactly two drank coffee is $3$ ways to select the one that does not times $0.81^2$ for the two that do times $0.19$ for the one that does not. Now do all three, which is easier and add.