For planes of a certain type, the actuarial estimate of the probability of a crash during a flight (including takeoff and landing) from New York to Chicago is $P_1$; from Chicago to L.A., it is $P_2$.
In experiment #1, a plane of that type is trying to fly from New York to L.A., with a stop in Chicago. Let $a$ denote the conditional probability that, if there is a crash, it occurs on the Chicago-L.A. leg of the journey.
In experiment #2, two different planes of the fatal type are involved; one is to fly from NY to Chicago, the other from Chicago to L.A. Let $b$ denote the conditional probability that the Chicago-L.A. plane crashed, if it is known that at least one of the two crashed, and let $c$ denote the conditional probability that the Chicago-L.A. plane crashed, if it is known that exactly one of the two crashed.
Express $a$, $b$, and $c$ in terms of $P_1$ and $P_2$, and show that $a$ ≤ $c$ ≤ $b$ for all possible $P_1, P_2$.
For part $a$, the probability of a crash would be $P_1 + P_2$, right?
Right now, for $a$ I have $a$ = $P_2/(P_1 + P_2)$
For $b$ I know that $P$(at least one) = 1 - $P$(no crash). So, would it be $P$($b$ | at least one crashed) = [ $P$($b$) $\cap$ $P$(at least one crashed) ] / $P$($b$)?
$a$ ≤ $c$ ≤ $b$ makes sense to me I think. The probability of $c$ would be 1/2, right? Given that one of the two planes crashed? And the probability of $b$ could be 1/2 (if only one plane crashed) or 1 if both crashed? $A$ has the smallest probability because it is not guaranteed that a crash even occurred.
I don’t know how to put the answers in terms of $P_1$ and $P_2$.
Be careful when we sum up probability, it might exceed $1$.
Probability of a crash can be computed by $1$ subtract away the probability of not crashing. $$1-(1-P_1)(1-P_2)=P_1+P_2-P_1P_2$$
Also, to have a crash at Chicago-LA route, we need to avoid crash in the first leg, $$a=\frac{(1-P_1)P_2}{P_1+P_2-P_1P_2}$$
Also, note that $b$ is a probability, not an event.
To compute $b$, we do not need to care about what happens in the first leg in the numerator.
$$b=\frac{P_2}{P_1+P_2-P_1P_2}$$
To compute $c$, the probability of exactly one crash is $$P_1(1-P_2)+P_2(1-P_1),$$
Hence $$c=\frac{P_2(1-P_1)}{P_1(1-P_2)+P_2(1-P_1)}$$
I will leave the comparison as an exercise.