Question

The probabilities of four mutually exclusive and exhaustive events A, B, C and D, satisfy the relation: 2P(A) = 3 P(B) = 4P(C) = 6P(D). The probability of event C is:
1. 3 / 14
2. 1 / 5
3. 4 / 15
4. 2 / 5

Answer 1

Correct Answer - Option 2 : 1 / 5

Given 

2P(A) = 3 P(B) = 4P(C) = 6P(D).

Formula used 

Probability = (possible outcomes/total outcomes)

Calculation 

Let total outcomes be k 

P(A) = k/2

P(B) = k/3

P(C) = k/4

p(D) = k/6

P(A) + P(B) + P(C) + P(D) = 1 

⇒ k/2 + k/3 + k/4 + k/6 = 1

⇒ k = 12/15 = 4/5

Probability of event C = (4/5)/(4) = 1/5

∴ The required answer is 1/5