Given two events A,B with P(A)=3/4 and P(B)=1/3, what is the smallest possible value of P(A∩B)? The largest? That is, find a and b such that,

a≤P(A∩B)≤b,

holds and any value in the closed interval [a,b] is possible.

a = 1/12

b = 1/3

0.34 1/3

To find the smallest possible value of P(A∩B), we can use the formula P(A∩B) = P(A) + P(B) - P(A∪B), where P(A∪B) is the probability of the union of events A and B.

Since P(A) = 3/4 and P(B) = 1/3, we need to find the smallest possible value of P(A∪B) in order to minimize P(A∩B).

To find the smallest value of P(A∪B), we can assume that events A and B are disjoint, meaning they have no elements in common. In this case, P(A∪B) = P(A) + P(B).

So, P(A∩B) will be smallest when A and B are disjoint, giving us the smallest possible value of P(A∩B) as 0.

Therefore, a = 0.

To find the largest possible value of P(A∩B), we need to find the largest possible value for the intersection of events A and B.

Since A and B are independent events, the intersection of A and B will be the product of their probabilities, P(A) * P(B).

Thus, P(A∩B) will be largest when A and B are independent, giving us the largest possible value of P(A∩B) as (3/4) * (1/3) = 1/4.

Therefore, b = 1/4.

Therefore, the smallest possible value of P(A∩B) is 0, and the largest possible value is 1/4.