For two events A and B, if P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. Then which of the following statements are TRUE ?
A. P(A∩B) = 0.3.
B. A and B are independent events.
C. P(A′∩B′) = 0.2.
D. A and B are mutually exclusive events.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A, B and C only
By the addition theorem:
\( \displaystyle P(A\cap B)=P(A)+P(B)-P(A\cup B) \)
\( \displaystyle =0.6+0.5-0.8=0.3 \)
\( \displaystyle P(A)\,P(B)=0.6\times 0.5=0.3 \)
By De Morgan’s law, \( A'\cap B'=(A\cup B)' \):
\( \displaystyle P(A'\cap B')=1-P(A\cup B) \)
\( \displaystyle =1-0.8=0.2 \)
So P(A∩B) = 0.3, P(A)P(B) = 0.3 and P(A′∩B′) = 0.2.
Checking the statements
(A) P(A∩B) = 0.3. — Correct. 0.6 + 0.5 − 0.8 = 0.3.
(B) A and B are independent events. — Correct. P(A)P(B) = 0.3 = P(A∩B).
(C) P(A′∩B′) = 0.2. — Correct. 1 − P(A ∪ B) = 1 − 0.8 = 0.2.
(D) A and B are mutually exclusive events. — Incorrect. Mutually exclusive events have P(A∩B) = 0, but here P(A∩B) = 0.3.
Tip: Independent events with non-zero probabilities can never be mutually exclusive, because then P(A∩B) = P(A)P(B) > 0.