Oranges are supplied from 3 parishes in Jamaica.C,M and A with probabilities P(C)=0.5,P(M)=0.1 and P(A)=0.4. Suppose that some oranges have a particular disease D such that P(D/C)=0.10,P(D/A)=0.25. What is the probability that if an orange selected at random from a combined supply is found to have the disease,it is from A?
If a diseased orange is found from the combined supply, there is a 2/3 chance that the orange is from A.
This is based on the following:
1. For C:
P(C) = 0.5 and P(D/C) = 0.1
Therefore, if an orange is picked at random from the supply, the chance that it is diseased and from "C" is:
0.5 * 0.1 = 0.05
2. Now for A:
P(A) = 0.4 and P(D/A) = 0.25
Again, if an orange is picked at random, the chance that it is diseased and from "A" is:
0.4 * 0.25 = 0.10
3. The chance of a diseased orange being selected is:
P(D) = 0.05 + 0.10 = 0.15
4. Therefore, if a diseased orange is found, the probability that it is from A is
P (A/D) = 0.10 / 0.15 = 2/3. Answer
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