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Muddy is caught with joint probability 1/15 at door one, 1/12 at door two and 1/6 at door three, giving 19/60 in all. Given that he is caught, what is the probability of door one or door two?
AIt is 9/19
BIt is 9/60
CIt is 10/19
DIt is 19/60
Answer & Solution
Correct answer: A. It is 9/19
1. Put the three joint probabilities over a common denominator of 60.
2. They become 4/60 for door one, 5/60 for door two and 10/60 for door three.
3. Doors one and two together account for 4/60 plus 5/60, which is 9/60.
4. Conditioning on being caught reduces the sample space to the total of 19/60.
5. Dividing 9/60 by 19/60 cancels the sixties and leaves 9/19.
6. Answering 9/60 forgets the conditioning and leaves the joint probability instead.
_Source: OpenStax Introductory Business Statistics (CC BY 4.0), Ch 3 "Probability Topics", section 3.4 Contingency Tables and Probability Trees_
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