Question

A random experiment has five outcomes \(w_1, w_2, w_3, w_4, w_5\). The probabilities of the occurrence of the outcomes \(w_1, w_2, w_4, w_5\) are respectively \( \frac{1}{6}, a, b, \frac{1}{12} \) such that \(12a + 12b - 1 = 0\). Then the probabilities of occurrence of the outcome \(w_3\) is:

• A. \( \frac{1}{3} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{1}{12} \)
• D. \( \frac{2}{3} \)

Hint:

In probability problems, always ensure that the sum of probabilities equals 1 and use the given conditions to solve for unknowns.

Answer 1

The Correct Option is D

We are given: \[ 12a + 12b - 1 = 0 \quad \text{and} \quad w_1 + w_2 + w_3 + w_4 + w_5 = 1 \] This means: \[ \frac{1}{6} + a + b + \frac{1}{12} = 1 \] To solve for \(a\) and \(b\), first, simplify the equation: \[ \frac{1}{6} + \frac{1}{12} + a + b = 1 \] \[ \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \] Thus, we have: \[ \frac{1}{4} + a + b = 1 \] \[ a + b = 1 - \frac{1}{4} = \frac{3}{4} \] Now, substitute the condition \(12a + 12b - 1 = 0\): \[ 12a + 12b = 1 \] \[ 12(a + b) = 1 \] \[ 12 \times \frac{3}{4} = 1 \quad \text{which is true}. \] Now, we know that the sum of all probabilities equals 1: \[ \frac{1}{6} + a + b + \frac{1}{12} + w_3 = 1 \] Substituting the values: \[ \frac{1}{6} + \frac{1}{12} + \frac{3}{4} + w_3 = 1 \] \[ w_3 = 1 - \left( \frac{1}{6} + \frac{1}{12} + \frac{3}{4} \right) \] \[ w_3 = 1 - \left( \frac{2}{12} + \frac{1}{12} + \frac{9}{12} \right) \] \[ w_3 = 1 - \frac{12}{12} = \frac{2}{3} \] Thus, the probability of occurrence of \(w_3\) is \( \frac{2}{3} \).