Answer to Question #32851 in Statistics and Probability for sandra
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)
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)
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
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).
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!