If a technician does not encounters any hardware problems, the time he requires to assemble a computer follows a normal distribution with a mean of $30$ minutes and a standard deviation of $3$ minutes. Let $T$ be the time in which he assembles a computer.
(a) Find the probability that it will take him more than $36$ minutes to assemble a computer given that he does not encounter hardware problems.
(b) When he encounters hardware problems the time to assemble a computer has a mean of 50 minutes and a standard deviation of $7$ minutes. Find the probability that it will take him more than $3$6 minutes to assemble a computer given that he encounters hardware problems.
(c) Suppose that he encounters a hardware problem $10\%$ of the time. If it took him more than $36$ minutes to assemble a computer, what is the probability that he encountered a hardware problem?
Let $Y$ be the event that the tech encounters a hardware problem.
For (a) I used $T\sim N(30,3)$ and I found $P(T>36\mid \overline Y) =1-P(T\leq36\mid \overline Y)=1-P(Z\leq2\mid \overline Y) = 1 - \Phi(2) = 0.0228$
Similarly for (b) I used $T\sim N(50,7)$ and found that $P(T>36\mid Y) = 0.9772$
Now for(c), I'm confuse as to how to set it up. Am I looking for a new random var?
I think is something like this:
How do I find $P(Y \mid T >36)$ How do I proceed?