$X$ ~ $exp (0.5)$ and $Y$ ~ $exp (1)$. X is the waiting time for the bus and Y is the waiting time for the taxi. Meaning half a bus is expected to arrive every 1 hour and 1 taxi is expected to arrive every 1 hour.
It is know that in $\frac{1}{3}$ of the days I take the taxi and in $\frac{2}{3}$ I take the bus. What is the probability that in a random day I took the bus, if it is known that I waited less than $1.2$ hours overall?
I first calculated $P(X < 1.2)$ but now I am not sure how to continue. How can I calculate the probability that.overall I waited less than $1.2$ hours? If I have that then I think I can get the solution
$\endgroup$ 41 Answer
$\begingroup$I think you are on the right track: the probability $P(X<Y|\min\{X,Y\}<1.2)$ is sought.
Set $A=\{\min\{X,Y\}<1.2\}$. By independence, $$P(A)=P(X<1.2,Y<1.2)=P(X<1.2)P(Y<1.2).$$ Further, by the Law of Total Probability (disintegration), $$P(\{X<Y\}\cap A)=\int_0^{1.2} P(X<y,X<1.2)dF_Y(y)=\int_0^{1.2} P(X<y)dF_Y(y), $$ where $F_Y(y)$ is the c.d.f. of $Y$. At this point, explicit calculation of $$P(X<Y|A)=\frac{P(\{X<Y\}\cap A)}{P(A)}$$ is readily available.
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