Question 1194705: Suppose P(A)=0.9, P(B)=0.4 and P(B|A)=0.2. What is P(A|B)?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Apply Bayes' Theorem to get...
P(A | B) = P(B | A)*P(A)/P(B)
P(A | B) = 0.2*0.9/0.4
P(A | B) = 0.18/0.4
P(A | B) = 0.45
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Another approach (slightly longer)
Recall the conditional probability P(A | B) is defined like such
P(A | B) = P(A and B)/P(B)
Similarly,
P(B | A) = P(A and B)/P(A)
Plug in the given values and solve for P(A and B)
P(B | A) = P(A and B)/P(A)
P(A and B) = P(B|A)*P(A)
P(A and B) = 0.2*0.9
P(A and B) = 0.18
Then we can say:
P(A | B) = P(A and B)/P(B)
P(A | B) = 0.18/0.4
P(A | B) = 0.45
As you can probably guess, Bayes' Theorem ties very closely to the conditional probability definition.
It shouldn't be too hard to prove that
P(A and B) = P(A | B)*P(B)
P(A and B) = P(B | A)*P(A)
based on the conditional probability definitions at the top of this section.
Because P(A and B) is equal to two different things, we can equate those right hand sides and say...
P(A | B)*P(B) = P(B | A)*P(A)
P(A | B) = P(B | A)*P(A)/P(B)
or we can say
P(A | B)*P(B) = P(B | A)*P(A)
P(B | A) = P(A | B)*P(B)/P(A)
Let me know if you need further help or clarification.
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Answer = 0.45
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