So I'm a math tutor and one of my students approached me with this problem:
A grocery store has six checkout lanes: four normal lanes, which are in use on average 80% of the time, and two express lanes, which are in use 60% of the time. If you choose to checkout at a randomly selected time, what is the probability that there is precisely one lane (of either kind) open?
Here is how I directed the student to its solution.
Suppose there are X normal lanes open and Y express lanes open at some randomly chosen time. IF we can assume that lanes are independently occupied (which is a stretch) then $ X \sim B(4,\frac{1}{5})$ and $Y \sim B(2,\frac{2}{5})$. We need to compute $P(X+Y=1).$ Using additivity and independence, we get$$ P(X+Y=1)=P(X=1)P(Y=0)+P(X=0)P(Y=1) \approx33.4\%$$
Evidently this problem was marked totally incorrect on this student's homework and I would like to know why. If anyone could help I'd be grateful. Thanks.