A company has two types of machine that run 24 hours a day, 7 days a week. Type A machine breaks down on average 5 breakdowns per day. This is equivalent to saying the mean wait time per breakdown is (1/5) days per breakdown. For Type B machine, the mean wait time between breakdowns is 0.35 days per breakdown. For each type of machine, assume breakdowns occur as a time homogeneous Poisson process, so that for each type of machine the waiting time for a breakdown has exponential distribution, but with differing parameters depending on whether the machine is type A or type B. Suppose 68% of the machines at this company are type A. You observe a machine but cannot tell what type it is. But you do observe that within the next 3 hours it does not break down. What is the posterior probability this is a type A machine? Answer as a decimal accurate to 4 decimal places.

Hint: Start by calculating the likelihood for each machine to not break down in 3 hours. This is equivalent to saying the waiting time to break down is more than 3 hours. This means you can use the EXPONENTIAL CDF to calculate these. The relevant calculations would have the form (>)=1−(,)=1−(1−−⋅) for type A machine
(>)=1−(,)=1−(1−−⋅) For type B machine
where λA and are the parameters. Make sure your parameters are correctly identified, and are either exact values, or if not, then accurate to 5 or more decimal places. (Be careful with what you choose as "t", as you notice that the problem specifies "day" not "hour" as the unit of measurement. ) Now make a two by two tree, where the first generation is whether a machine is type A or type B, and the second generation is whether the machine does not break down in 3 hours, or else does. The final answer is the probability of the machine being type A GIVEN evidence that the machine does not break down in 3 hours. In other words, it is a posterior probability.