Question

If A, B, C are three mutually exclusive and exhaustive events of an experiment such that P(A) = 2P(B) = 3P(C),then P(B) is equal to

• A. \(\frac{1}{11}\)
• B. \(\frac{2}{11}\)
• C. \(\frac{3}{11}\)
• D. \(\frac{4}{11}\)

Answer 1

The Correct Option is C

Given that A, B, and C are three mutually exclusive and exhaustive events of an experiment, such that \(P(A) = 2P(B) = 3P(C)\). 

We need to find \(P(B)\).

Since A, B, and C are mutually exclusive and exhaustive, we have:

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

Let \(P(B) = x\). Then, \(P(A) = 2x\) and \(2x = 3P(C) \implies P(C) = \frac{2}{3}x\).

Substituting these into the equation \(P(A) + P(B) + P(C) = 1\), we get:

\(2x + x + \frac{2}{3}x = 1\)

\(3x + \frac{2}{3}x = 1\)

\(\frac{9x + 2x}{3} = 1\)

\(\frac{11x}{3} = 1\)

\(11x = 3\)

\(x = \frac{3}{11}\)

Therefore, \(P(B) = \frac{3}{11}\).

Thus, the correct option is (C) \(\frac{3}{11}\).

Answer 2

Let \( P(A) \), \( P(B) \), \( P(C) \) be the probabilities of events \( A \), \( B \), \( C \). 

Since \( A \), \( B \), \( C \) are mutually exclusive and exhaustive, \( P(A) + P(B) + P(C) = 1 \). 

Given \( P(A) = 2P(B) = 3P(C) \), then \( P(B) = \frac{P(A)}{2} \) and \( P(C) = \frac{P(A)}{3} \). 

Substituting into the equation: \[ P(A) + \frac{P(A)}{2} + \frac{P(A)}{3} = 1 \] Multiply through by 6 to eliminate the fractions: \[ 6P(A) + 3P(A) + 2P(A) = 6 \] Simplify: \[ 11P(A) = 6 \] Solve for \( P(A) \): \[ P(A) = \frac{6}{11} \] Now calculate \( P(B) \): \[ P(B) = \frac{P(A)}{2} = \frac{\frac{6}{11}}{2} = \frac{3}{11} \] Final Answer: The final answer is \( \boxed{\frac{3}{11}} \).