SOLVED - exercise involving the exponential distribution and conditional probabiliy.

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I have a box with two types of elements A and B. There are 7 elements in total. 1 of type A, and 6 of type B.

Both types of elements have some property Z that's exponentially distributed.

The expectation value of Z for type A is 11

The expectation value of Z for type B is 18.5

I pick an element from the box at random.

Given that Z for the element I've picked is 15 or greater, what is the probability that I picked an element of type B?

-- I can do this problem "in reverse", e.g.:

  • Given that I pick an element of type B the probability that Z is greater than 15 is 0.444.

  • Given that I pick an element of type A the probability that Z is greater than 15 is 0.255.

but I don't understand how to proceed with the stated problem.

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This looks like a Bayes theorem problem:

$$ P(B | Z > 15) = \frac{P(B) P(Z>15 | B)}{P(Z>15)} = \frac{P(B) P(Z>15 | B)}{P(A)P(Z>15 | A) + P(B)P(Z>15 | B)} .$$

You know that $P(A)$, $P(B)$ are $1/7$ and $6/7$, and you've already calculated $$P(Z>15|A) = 0.255, $$ $$ P(Z>15 | B) = 0.444 \quad .$$