Question

Let \( A, B, C \) be three mutually and exhaustive events of an experiment. If \( 2P(A) = 3P(B) = 4P(C) \), then \( P(C) \) is equal to:

• A. \( \frac{3}{13} \)
• B. \( \frac{4}{13} \)
• C. \( \frac{5}{13} \)
• D. \( \frac{6}{13} \)
• E. \( \frac{7}{13} \)

Hint:

When dealing with mutually exclusive and exhaustive events, use the total probability formula \( P(A) + P(B) + P(C) = 1 \) and express each probability in terms of a single variable to simplify the problem.

Answer 1

The Correct Option is A

We are given that \( 2P(A) = 3P(B) = 4P(C) \), and that \( A, B, C \) are mutually exclusive and exhaustive events. 
This means:

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

Let:

\[ P(A) = x, \quad P(B) = y, \quad P(C) = z \]

From the given relationship:

\[ 2x = 3y = 4z \]

Thus, we can express \( y \) and \( z \) in terms of \( x \):

\[ y = \frac{2}{3}x \quad \text{and} \quad z = \frac{2}{4}x = \frac{1}{2}x \]

Now, substitute these expressions for \( y \) and \( z \) into the equation \( P(A) + P(B) + P(C) = 1 \):

\[ x + \frac{2}{3}x + \frac{1}{2}x = 1 \]

To solve this, first find a common denominator for the terms:

\[ \frac{6}{6}x + \frac{4}{6}x + \frac{3}{6}x = 1 \]

\[ \frac{13}{6}x = 1 \]

\[ x = \frac{6}{13} \]

Thus:

\[ P(C) = z = \frac{1}{2}x = \frac{1}{2} \times \frac{6}{13} = \frac{3}{13} \]

Thus, the correct answer is option (A), \( P(C) = \frac{3}{13} \).