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# Answer to Question #32851 in Statistics and Probability for sandra

Question #32851
suppose it is estimated that a person visiting Florida will visit Disney World, Busch Gardens or both with probabilities 0.7, 0.1, and 0.2 respectively. Find the probability that a person visiting Florida will visit Busch Gardens given that the person did not visit DIsney World.
Let A is event that person visiting Florida will visit Disney World, P(A) = 0.7, P(not_A) = 1 - 0.7 = 0.3
Let B is event that person visiting Florida will visit Busch Gardens, P(B) = 0.1.
Then A and B is event that person visiting Florida will visit Disney World and Busch Gardens, P(A and B) = 0.2.

Let not_D is opposition for event D.
Let's find P(B|not_A)

Solution
First of all in condition we have error. Not for all numbers there is solution.

We use condition probability:
P(B|not_A) = P(B and not_A)/P(not_A)

We use Inclusion–exclusion principle:
P((B or not_A) and A) = P((B and A) or (not_A and A)) = P(B and A)

Also
P((B or not_A) and A) = P(B or not_A) + P(A) - P(B or not_A or A) = P(B or not_A) + P(A) - 1.
We have P(B or not_A) = P(B and A) - P(A) + 1 = 0.2 - 0.7 + 1 = 0.5

Then
P(B and not_A) = P(B) + P(not_A) - P(B or not_A) = 0.1 + 0.3 - 0.5 = -0.1
but it makes no sense becase probability >= 0 always and <= 1.

This happened because P(A and B) > P(B). A and B are included in B that's why always we have P(A and B) <= P(B).
If we have P(B) = 0.3 then
P(B and not_A) = 0.3 + 0.3 - 0.5 = 0.1.
And P(B|not_A) = 0.1/0.3 = 0.33(3).

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