Question

If \( \frac{1 + 3p}{4}, \frac{1 - p}{3}, \frac{1 - 3p}{2} \) are the probabilities of three mutually exclusive and exhaustive events, then the value of \( p \) is:

• A. \( \frac{1}{3} \)
• B. \( \frac{12}{13} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{1}{13} \)
• E. \( \frac{2}{13} \)

Hint:

For mutually exclusive and exhaustive events, the sum of their probabilities must always equal 1. Use this condition to form an equation and solve for the unknown.

Answer 1

The Correct Option is D

Since the events are mutually exclusive and exhaustive, the sum of their probabilities must be 1. 
Therefore, we have the equation: \[ \frac{1 + 3p}{4} + \frac{1 - p}{3} + \frac{1 - 3p}{2} = 1 \] 
To solve this equation, we first take the least common denominator (LCD) of 4, 3, and 2, which is 12. 
We rewrite the equation as: \[ \frac{3(1 + 3p)}{12} + \frac{4(1 - p)}{12} + \frac{6(1 - 3p)}{12} = 1 \] 
Now, multiply both sides of the equation by 12: \[ 3(1 + 3p) + 4(1 - p) + 6(1 - 3p) = 12 \] Expanding the terms: \[ 3 + 9p + 4 - 4p + 6 - 18p = 12 \]
Simplifying: \[ 3 + 4 + 6 + (9p - 4p - 18p) = 12 \] \[ 13 - 13p = 12 \] \[ -13p = -1 \quad \Rightarrow \quad p = \frac{1}{13} \]