Machine 1 is currently working. Machine 2 will be put in use at time t from now. If the lifeti

Misael Matthews

Misael Matthews

Answered question

2022-06-22

Machine 1 is currently working. Machine 2 will be put in use at time t from now. If the lifetime of machine i is exponential with rate λ i , i = 1 , 2 , what is the probability that machine 1 is the first machine to fail?

The solution is as follows:
P ( M 1 < M 2 ) = P ( M 1 < M 2 | M 1 < t ) P ( M 1 < t ) + P ( M 1 < M 2 | M 1 > t ) P ( M 1 > t ) .
Now, I understand this solution. It states, in light of Bayes's Rule*, that we desire the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is less than t in addition to the probability that the lifetime of machine 1 is less than the lifetime of machine 2 and the lifetime of machine 1 is greater than t. That all makes sense to me. My question, however, is why is it not simply
P ( M 1 < M 2 ) = P ( M 1 < t ) + P ( M 1 < M 2 | M 1 > t ) .
Linguistically, this seems to satisfy the solution to the problem; i.e., "the probability that the lifetime of machine 1 is less than t, in addition to the probability that the lifetime of machine 1 is less than that of machine 2 given that the lifetime of machine 1 is greater than t. I understand that the first term here is actually the same as the solution's, and that the solution formulates it as it does for illustrative purposes; but the second term certainly has a different value than that of the solution's.

I hope my question here makes sense. If any further clarification is needed, please let me know. Also, I understand that this exact problem appears elsewhere on stackexchange, but it was more about the computation.

*The Bayes's Rule to which I'm referring is, of course, the following formulation:
P ( A B ) = P ( A | B ) P ( B )

Answer & Explanation

Christina Ward

Christina Ward

Beginner2022-06-23Added 19 answers

There's no justification for the "given that the lifetime of machine 1 is greater than t" part. This is not given. It should be and, as in your first paraphrase. If P ( M 1 > t ) is very low, the contribution from that term should be very low; whereas in your version its contribution could be up to 1 no matter how unlikely M 1 > t is.

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