Question 1125789: Let E, F and G be three events in S with P(E) = 0.65, P(F) = 0.57, P(G) = 0.55, P(E ∩ F) = 0.35, P(E ∩ G) = 0.39, P(F ∩ G) = 0.4, and P(E ∩ F ∩ G) = 0.3. Find P(EC ∪ FC ∪ GC).
P(EC ∪ FC ∪ GC) =
Found 2 solutions by kasperk, ikleyn:
Answer by kasperk(4)   (Show Source):
You can put this solution on YOUR website! For this explanation, please refer to the Venn diagram completed below with sections A, B, C, D, E, F and G.
We start by filling in section A, which represents the intersection of all three events, P(E ∩ F ∩ G) = 0.3.
Then we go to each of the other intersections of exactly two events. Since section A = 0.3 is part of the sections representing E∩F, E∩G and F∩G, we must subtract 0.3 from each of those double intersections, thereby getting:
Section B = P(E∩F) = 0.35-.3 = .05
Section C = P(E∩G) = 0.39-.3 = .09
Section D = P(F∩G) = 0.4-.3 = .1
Now you fill in the remaining sections, which is the necessary difference to make each full circle add up to P(E), P(F) and P(G).
Section E = P(E) - (.05 + .3 + .09) = .65 - .44 = .21
Section F = P(F) - (.05 + .3 + .1) = .57 - .45 = .12
Section G = P(G) - (.3 + .09 + .1) = .55 - .49 = .06
Now add up every number in the 7 sections =
P(E ∪ F ∪ G) = .93
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
Regarding the post by @kasperk, I'd like to notice that it is IRRELEVANT to the posed question.
The post by @kasperk calculates P(E U F U G), while the question asks about P(EC U FC U GC).
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