Exponential Distribution - Probabilities I'm trying to figure out the answer to the following question on a past exam given for practice. Since there are no solutions, I was hoping I could get the help needed to figure it out.

A BS1300 big-screen TV may die due to screen failure or power supply failure.
The time until power supply failure is exponential with parameter 0.00001.
The time until screen failure is exponential with parameter 0.00002. The time is in hours.
a) Which failure is more probable to occur first?
For this problem, I simply got the expected values and whichever was lower, it should be more probable as the time expected for it to fail was lower.
$E(X) = 100000$ hours for power supply, $E(X) = 50000$ hours for screen failure. Therefore, screen is more likely to fail first.
b) What is the probability that a BS1300 will die within 10000 hours?
PS = power supply, SF = screen failure
$P(PS = 10000) = 0.000016374 P(SF = 10000) = 0.000025422$. If we add it up, we get PS or SF $= 0.000025422$.
c) If the BS1300 is known to be dead, what is the probability that its power supply failed?
Part c), I'm not sure how to figure this out. I tried drawing a venn diagram but I'm unsure how to do it for exponential. Help would be appreciated on how to solve this problem.
 A: Hint :  Here is the edited solution:
Ist part is correct
2nd Part
x1 - time until failure happens with power supply
x2 - time until failure happens with screen failure
Define x1 be the R.V with pdf = f(x1) = $\lambda_1$*$e^{-\lambda_1*x1}$
Define x2 be the R.V with pdf = f(x2) = $\lambda_2$*$e^{-\lambda_2*x2}$
P(X1<=10000) = $1-e^{-\lambda_1*x1}$ ............(1)
P(X2<=10000) = $1-e^{-\lambda_2*x2}$ ............(2)
Computing these two we get 
(1) = 0.09516
(2) = 0.18126
Probability that the machine will fail = Max(0.09516,0.18126)
Another property of exponential distribution that is derived  is
If X1 and X2 are independent exponential RVs with parameters λ1 and λ2,
P(X1 < X2) = $\frac{λ_1}{λ_1 + λ_2}$
P(X1 < X2) essentially encapsulates the fact that the time until failure due to power supply precedes the time until failure due to Screen Failure is indeed the probability that the failure is due to power supply.
MoonKnight:  I thought about your comment and realized that two CDFs will approach 1 as move to the extreme right on time. So I refrain from using that argument.  But I would have liked to see you completed solution with your reasoning.
Thanks
Satish
A: (a) correct
(b) wrong, first of all here you should use the CDF but not the PDF. Also you cannot simply add them up.
PG for power good, SG for screen good.
$P(PG)=1-p(PS)$, $P(SG)=1-P(SF)$. 
BS1300 is not dead, equivalent to both PG and SG. Since PG and SG are independent.
So $P(PG\cap SG)=P(PG)P(SG)=...$
