Question 588334: Consider two events A and B with P(A) = 8/10 and P(B) = 9/10.
Prove that P(A|B) greater than or equal 7/9.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! P(A and B) = P(A) + P(B) - P(A or B)
P(A and B) = 8/10 + 9/10 - x
P(A and B) = 17/10 - x
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P(A | B) = P(A and B)/P(B)
P(A | B) = (17/10 - x)/(9/10)
Since 0 < P(A or B) < 1, this means that the smallest value of x is x = 0 and the largest value of P(A | B) is P(A | B) = (17/10 - 0)/(9/10) = 17/9
and the largest value is x = 1, so the smallest value of P(A | B) = (17/10 - x)/(9/10) is P(A | B) = (17/10 - 1)/(9/10) = 7/9
Note: the largest and smallest are mixed up because if you have something like 1-x (where 0 < x < 1), then as x increases 1-x decreases (and vice versa). Similar reasoning applies when x decreases.
Therefore, if P(A) = 8/10 and P(B) = 9/10, then P(A|B) is greater than or equal to 7/9.
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