Question

Let \(A\) and \(B\) be two events such that one of the two events must occur. Given that the chance of occurrence of \(A\) is \( \frac{2}{3} \) the chance of occurrence of \(B\), then odds in favour of \(B\) is

• A. \(3:5\)
• B. \(2:5\)
• C. \(3:2\)
• D. \(2:3\)

Hint:

Odds in favour of an event = Probability of event : Probability of its complement.

Answer 1

The Correct Option is C

Step 1: Understand the condition “one must occur”.
This means events \(A\) and \(B\) are complementary in nature:
\[ P(A) + P(B) = 1. \]

Step 2: Use the given relation.
It is given:
\[ P(A) = \frac{2}{3}P(B). \]

Step 3: Substitute in the probability equation.
\[ \frac{2}{3}P(B) + P(B) = 1. \]
\[ \frac{2}{3}P(B) + \frac{3}{3}P(B) = 1. \]
\[ \frac{5}{3}P(B) = 1. \]

Step 4: Solve for \(P(B)\).
\[ P(B) = \frac{3}{5}. \]

Step 5: Find probability of not \(B\).
\[ P(B') = 1 - P(B) = 1 - \frac{3}{5} = \frac{2}{5}. \]

Step 6: Write odds in favour of \(B\).
Odds in favour = \( P(B) : P(B') \)
\[ = \frac{3}{5} : \frac{2}{5}. \]

Step 7: Simplify the ratio.
\[ = 3 : 2. \]
Final Answer:
\[ \boxed{3:2}. \]