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OleMash [197]
4 years ago
7

If the probability of A given B is 0.8 and the probability of A is 0.12 what is the probability of B

Mathematics
1 answer:
horsena [70]4 years ago
5 0

We can't tell from the information given. What we do know is

P(A ∩ B) = P(A|B) P(B)

We're given the conditional probability but not the probability of the conjunction.

If we were told P(A|B)=0.8 and P(A ∩ B)=0.12 we could conclude

P(B)=P(A ∩ B)/P(A|B) = 0.12/0.8 = 0.15

but that wasn't the problem asked.



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Step 1; Assume that the probability of the first inspector missing a defective part is P(A) and the probability of the second inspector missing those that do get past the first inspector is P(B).

Step 2; It is given that P(A) = 0.1, we convert this into a fraction so that the final probability will be a fraction and not a decimal.

P(A) = 0.1 = \frac{1}{10}.

It is given that the second inspector misses 5 out of 10 that get past the first inspector, so P(B) = \frac{5}{10}.

Step 3; To calculate the probability of both inspectors missing a defective part, we multiply both the probabilities.

P(A and B happening) = P(A) × P(B) = \frac{1}{10} × \frac{5}{10} = \frac{5}{100} = \frac{1}{20} = 0.05%. So there is a 0.05% chance of both inspectors missing a defective part.

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