Probabilities

If probabilities of A, B and A ∩ B are P (A) = 0.62 P (B) = 0.78 and P (A ∩ B) = 0.26 calculate the following probability (of union. intersect and opposite and its combinations):

Result

P(A′) =  0.38
P(B′) =  0.22
P(A ∪ B) =  1.14
P(A′∩ B) =  0.52
P(A ∩ B′) =  0.36
P[( A ∪ B)′] =  -0.14
P( A′ ∪ B) =  0.64

Solution:

P(A′) = 1-0.62 = 0.38
P(B′) = 1-0.78 = 0.22
P(A ∪ B) = 0.62+0.78-0.26 = 1.14
P(A′∩ B) = 0.78-0.26 = 0.52
P(A ∩ B′) = 0.62-0.26 = 0.36
P[( A ∪ B)′] = 1-(0.62+0.78-0.26) = -0.14
P( A′ ∪ B) = 1-0.62+ 0.26 = 0.64







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To solve this example are needed these knowledge from mathematics:

Would you like to compute count of combinations?

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