Popcorn Probability Question:
The time to microwave a bag of popcorn using the automatic setting can be treated as a random variable having a normal distribution with mean 2 min and standard deviation 15 seconds. 
three independent bags of popcorn A,B and C are selected. The two bags A and B are put into a microwave immediately. As soon as bag A pops, the bag C is put into microwave. Find the probability that B pops before C.
I really don't know where do start. I think it would be something along the lines of P(B pops before A)*P(B pops before c - 2). Any help would be greatly appreciated,
 thanks.
 A: We are looking for
$$P(A+C-B)>0$$
where $A,B,C$ are independent and normally distributed with mean 120 and variance $15^2=225$. By the properties of linear combinations of normal distributions we find that $A+C-B$ is normally distributed with mean 120 and variance $225×3=675$ (or a standard deviation of $\sqrt{675}$). The z-score of 0 on this normal distribution is $\frac{0-120}{\sqrt{675}}=-4.62$.
Putting it all together, we have
$$P(A+C-B>0)=P(\mathcal N(120,675)>0)=P(Z>-4.62)=0.999998.$$
A: Let $X$, $Y$ and $Z$ be the times to microwave bags $A$, $B$, and $C$ respectively. I assume that $X,Y,Z$ are the same as the times to "pop" these bags.
According to the design of the experiment, bag $C$ will pop after $X+Z$ minutes, and bag $B$ will pop after $Y$ minutes. You want the probability that $Y<X+Z$. To compute this probability, note that $D:=X+Z-Y$ is a linear combination of independent normal variables, so $D$ has a normal distribution. What are its mean and variance? Argue that the desired probability is $P(D>0)$. 
