Would anyone be able to show me how to figure out the following question? I'm not particularly interested in the answer, more so in the steps towards obtaining it. Thank you.
The manager of a fast food restaurant knows from experience that 35% of customers who order a meal will also order coffee. There are 20 customers at the restaurant.
What is the most likely number of customers who will order coffee?
Let $X$ be the number of coffee-orderers. Assuming (unreasonably) independence, and that each of the $20$ people orders coffee with probability $0.35$, the random variable $X$ has binomial distribution, and $$\Pr(X=k)=\binom{20}{k}(0.35)^k (0.65)^{20-k}.$$
It turns out that as $k$ increases from $0$, the probability increases and then decreases, reaching a maximum not far from $(0.35)(20)$.
Compute until you find the answer. Note that there are computational shortcuts: if you have found $\binom{20}{k}(0.35)^k(0.65)^{20-k}$, you can compute the "next term" relatively simply. For $\binom{n}{k+1}=\frac{n-k}{k+1}\binom{n}{k}$.
Remark: There is general theory about the mode of a binomial, and it has been done in MSE, at least once by me. But searching is not particularly easy.
Added: Semi-miraculously, I found one of the places where general theory is developed.
