Question

Let A and B be two events such that \(\rm P\left( \overline{A\cup B} \right)=\frac{1}{6}\), \(\rm P\left( A\cap B \right)=\frac{1}{4}\) and \(\rm P\left( \overline{A} \right)=\frac{1}{4}\), where A̅ is the complementary event of A. Then A and B are:

1. Equally likely but not independent.
2. Equally likely and mutually exclusive.
3. Mutually exclusive and independent.
4. Independent but not equally likely.

Answer 1

Correct Answer - Option 4 : Independent but not equally likely.

Concept:

• If P(A) = P(B), then the events are said to be equally likely.
• If A and B are independent events, then P(A ∩ B) = P(A) × P(B).
• If A and B are mutually exclusive events, then P(A ∩ B) = 0.
• For two events A and B, we have P(A ∪ B) = P(A) + P(B) - P(A ∩ B).
• P(E̅) = 1 - P(E), where E̅ is the complementary event of E.

 

Calculation:

Using the definition of complementary events:

P(A ∪ B) = \(\rm 1-P\left( \overline{A\cup B} \right)=1-\frac{1}{6}=\frac{5}{6}\)

P(A) = 1 - P(A̅) = \(\rm 1-\frac{1}{4}=\frac{3}{4}\)

Let us find out P(B) and examine the values of P(A ∩ B) and P(A) × P(B) to determine whether the events are equally likely and mutually exclusive or independent.

Using P(A ∪ B) = P(A) + P(B) - P(A ∩ B), we get:

\(\rm \frac{5}{6}=\frac{3}{4}+P\left( B \right)-\frac{1}{4}\)

⇒ P(B) = \(\rm \frac{5}{6}-\frac{3}{4}+\frac{1}{4}=\frac{1}{3}\)

Since, P(A) ≠ P(B), the events are not equally likely.

Also,P(A) × P(B) = \(\rm \frac{3}{4}× \frac{1}{3}=\frac{1}{4}\) = P(A ∩ B), so the events are independent.

The correct answer option is D. Independent but not equally likely.