Answer:
(a) max P(A∩B) = 1/3; min P(A∩B) = 1/12
(b) max P(A∪B) = 1; min P(A∪B) = 3/4
Explanation:
Let the universal set be the numbers 1–12, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, each with probability 1/12.
Let event A be any of the numbers 1–9, {1, 2, 3, 4, 5, 6, 7, 8, 9}. If a number is chosen at random from U, the probability of event A is 9/12 = 3/4.
a1) Let event B be any of the numbers 1–4, {1, 2, 3, 4}. If a number is chosen at random from U, the probability of event B is 4/12 = 1/3.
The set A∩B is the numbers 1–4, {1, 2, 3, 4}, so the probability of that event is also 4/12 = 1/3.
In general the maximum value of P(A∩B) will be min(P(A), P(B)). Here, that is min(3/4, 1/3) = 1/3.
__
a2) Let event B be any of the numbers 9–12, {9, 10, 11, 12}. If a number is chosen at random from U, the probability of event B is 4/12 = 1/3. The set A∩B is the number {9}, so the probability of that event is 1/12.
In general, the minimum value of P(A∩B) is max(0, P(A) +P(B) -1). Here, that is max(0, 3/4 +1/3 -1) = 1/12.
__
b1) Let event B be defined as in (a2), the numbers 9–12. Then A∪B is the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}, which is equal to the universal set, U. That is, the probability of event A∪B when drawing a number from U is 1.
In general, the maximum value of P(A∪B) is min(1, P(A)+P(B)). Here, that is min(1, 3/4+1/3) = 1.
__
b2) Let event B be defined as in (a1), the numbers 1–4. Then A∪B is the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. If a number is chosen at random from U, the probability of event A∪B is 9/12 = 3/4.
In general, the minimum value of P(A∪B) is max(P(A), P(B)). Here, that is max(3/4, 1/3) = 3/4.