Financial Risk Manager Part 1
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A fruit juice shop allows customers to choose apple juice, mango juice or passion juice. The probability of a customer ordering passion juice is 0.45, mango juice and apple juice 0.19, passion juice and mango juice 0.15, passion juice and apple juice 0.25, passion juice or mango juice 0.6, passion juice or apple juice 0.84, and 0.9 for at least one of them. Find the probability that a customer orders all the three juices.
Explanation:
Explanation
Let's define the events:
- P = Passion juice
- M = Mango juice
- A = Apple juice
Given probabilities:
- P(P) = 0.45
- P(M ∩ A) = 0.19
- P(P ∩ M) = 0.15
- P(P ∩ A) = 0.25
- P(P ∪ M) = 0.6
- P(P ∪ A) = 0.84
- P(P ∪ M ∪ A) = 0.9
We need to find P(P ∩ M ∩ A).
Using the inclusion-exclusion principle for three events: P(P ∪ M ∪ A) = P(P) + P(M) + P(A) - P(P∩M) - P(P∩A) - P(M∩A) + P(P∩M∩A)
We know P(P ∪ M ∪ A) = 0.9
We need to find P(M) and P(A).
From P(P ∪ M) = P(P) + P(M) - P(P∩M) = 0.6 0.6 = 0.45 + P(M) - 0.15 0.6 = 0.3 + P(M) P(M) = 0.3
From P(P ∪ A) = P(P) + P(A) - P(P∩A) = 0.84 0.84 = 0.45 + P(A) - 0.25 0.84 = 0.2 + P(A) P(A) = 0.64
Now plug into the inclusion-exclusion formula: 0.9 = 0.45 + 0.3 + 0.64 - 0.15 - 0.25 - 0.19 + P(P∩M∩A) 0.9 = 1.39 - 0.59 + P(P∩M∩A) 0.9 = 0.8 + P(P∩M∩A) P(P∩M∩A) = 0.9 - 0.8 = 0.1
Therefore, the probability that a customer orders all three juices is 0.1.