Question

One of the two events must occur. The chance of one is \( \frac{2}{3} \) of the other, then odds in favour of the other are

• A. 2 : 3
• B. 1 : 3
• C. 3 : 1
• D. 3 : 2

Hint:

When two events must occur and one is a multiple of the other, use total probability = 1 and solve algebraically.

Answer 1

The Correct Option is D

Let the probability of the first event be \( P_1 \), and of the second event be \( P_2 \). We are told: \[ P_1 = \frac{2}{3} P_2 \quad \text{and} \quad P_1 + P_2 = 1 \] Substitute: \[ \frac{2}{3}P_2 + P_2 = 1 \Rightarrow \frac{5}{3}P_2 = 1 \Rightarrow P_2 = \frac{3}{5} \Rightarrow P_1 = \frac{2}{5} \] Odds in favour of the second event = \[ \frac{P_2}{1 - P_2} = \frac{3/5}{2/5} = \frac{3}{2} \] Hence, the odds are 3:2.