Question

A, B, and C are three mutually exclusive and exhaustive events associated with a random experiment. If P(B) = \(\rm \frac{3}{2}P(A)\) and P(C) = \(\rm \frac{1}{2}P(B)\) then value of P(A) is:
1. \(\dfrac{1}{13}\)
2. \(\dfrac{2}{13}\)
3. \(\dfrac{4}{13}\)
4. \(\dfrac{3}{13}\)

Answer 1

Correct Answer - Option 3 : \(\dfrac{4}{13}\)

Concept:

A, B, and C are mutually exclusive events,

P(A) + P(B) + P(C) = 1

Calculation:

P(B) = \(\rm \frac{3}{2}P(A)\) and P(C) = \(\rm \frac{1}{2}P(B)\)

⇒ Let P(A) = p

⇒ P(B) = \(\rm \frac{3p}{2}\) and P(C) = \(\rm \frac{3p}{4}\)

Since A, B, C are mutually exclusive and exhaustive events associated with a random experiment.

⇒ A ∪ B ∪ C = S

⇒ P(A ∪ B ∪ C) = P(S)

⇒ P(A ∪ B ∪ C) = 1

⇒ P(A) + P(B) + P(C) = 1

⇒ p + \(\rm \frac{3p}{2}\) + \(\rm \frac{3p}{4}\) = 1

⇒ p = \(\rm \dfrac{4}{13}\)

∴ P(A) = \(\rm \dfrac{4}{13}\)