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If A and B ore two events such that P(A)=(1)/(4),P(B)=(1)/(2) and P(A and B)>(1)/(8) Find

(i) P(A or B)
(ii) P(notA and not B)

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Final answer:

To find P(A or B) and P(not A and not B), use the formulas P(A or B) = P(A) + P(B) - P(A and B) and P(not A and not B) = 1 - P(A) - P(B) + P(A and B).

Step-by-step explanation:

To find the probability of the union of two events (A or B), we use the formula: P(A or B) = P(A) + P(B) - P(A and B). Given that P(A) = 1/4, P(B) = 1/2, and P(A and B) > 1/8, we can substitute these values into the formula to find the probabilities.

(i) P(A or B) = P(A) + P(B) - P(A and B)

Plugging in the values, P(A or B) = 1/4 + 1/2 - P(A and B) > 1/8

Using the given information that P(A and B) > 1/8, we know that the probability of A and B happening together is greater than 1/8.

(ii) P(not A and not B) = 1 - P(A) - P(B) + P(A and B)

Substituting the given values, P(not A and not B) = 1 - 1/4 - 1/2 + P(A and B)

Since we don't have the exact value of P(A and B), we can't calculate P(not A and not B) accurately without that information.

User Oleg Gordiichuk
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