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$X$ ~ $exp (0.5)$ 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

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    If I understand the question correctly, you're to compute $\Pr(X – 2017-01-14
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    Why? Can you explain please? – 2017-01-14
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    You take the bus if it arrives first (i.e. $X – 2017-01-14
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    @yurnero : but $X < 1.2$ is not the same event than "waiting less than 1.2 hours". I would rather calculate $P( X < Y | \min (X,Y) < 1.2 )$ – 2017-01-14

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I think you are on the right track: the probability $P(X

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