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Therea are two approaches, leading to the same solution
P(A or B or C) = P (A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P( A and B and C)
For proof, see:
https://math.stackexchange.com/questions/1291012/how-to-proof-formula-for-general-addition-rule-of-three-events
or obviously simpler, as most people here seem to have done : 1 - P(not A)*P(not B)*P(not C)
I am glad some used the generalized form of solving this.
I'm a bit confused as we are given probablity for only events A, B &C only so how will you calculate P(A∩B), P(A∩C), P(B∩C), P(A∩B∩C).
Since they are independent events: P(A∩B) = P(A)*P(B)
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Basic Probability Puzzles #9
You are viewing a single comment's thread. Return to all comments →
Therea are two approaches, leading to the same solution
P(A or B or C) = P (A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) + P( A and B and C)
For proof, see:
https://math.stackexchange.com/questions/1291012/how-to-proof-formula-for-general-addition-rule-of-three-events
or obviously simpler, as most people here seem to have done : 1 - P(not A)*P(not B)*P(not C)
I am glad some used the generalized form of solving this.
I'm a bit confused as we are given probablity for only events A, B &C only so how will you calculate P(A∩B), P(A∩C), P(B∩C), P(A∩B∩C).
Since they are independent events: P(A∩B) = P(A)*P(B)