A national polling company polled Canadians about their coffee consumption and made the following discovery: $85$% 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
Part (a) both consume at least one cup of coffee on the chosen day?
Part (b) Refer to the two Canadians chosen in part (a). Find the probability that neither consumes any coffee over the course of the chosen day.
Part (c) 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.
Part (d) Suppose you are to randomly pick $x$-Canadians until the probability that at least one of them has consumed at least one cup of coffee (on the randomly chosen day) is at least $0.96$. How large must $x$ be?
I'm completely lost on this question, would very much appreciate some help.
Best Answer
Since the question isn't clear about the distinction between "random day" and "every day" is, assume $85%$ drink coffee and $15%$ don't.
a) Given two Canadians $A,B$, assume the events $A$ drinks coffee and $B$ drinks coffee are independent. So the answer is $.85*.85$.
b) By same logic, $.15*.15$.
c) The probability that all three drink coffee is $.85^3$. The probability that two drink and the third doesn't is $.85^2*.15$, however there are 3 ways this can happen (pick one of three to not drink) so multiply this by 3. The answer is $.85^3+3(.85^2\cdot0.15)$
d) Find the opposite: what's the probability that of $n$ people, all $n$ of them don't drink coffee?