Question

Three students $ S_1, S_2, $ and $ S_3 $ are given a problem to solve. Consider the following events: $ U $: At least one of $ S_1, S_2, S_3 $ can solve the problem,
$ V $: $ S_1 $ can solve the problem, given that neither $ S_2 $ nor $ S_3 $ can solve the problem,
$ W $: $ S_2 $ can solve the problem and $ S_3 $ cannot solve the problem,
$ T $: $ S_3 $ can solve the problem. For any event $ E $, let $ P(E) $ denote the probability of $ E $. If $$ P(U) = \frac{1}{2}, \quad P(V) = \frac{1}{10}, \quad \text{and} \quad P(W) = \frac{1}{12}, $$ then $ P(T) $ is equal to:

• A. \( \dfrac{13}{36} \)
• B. \( \dfrac{1}{3} \)
• C. \( \dfrac{19}{60} \)
• D. \( \dfrac{1}{4} \)

Hint:

To handle composite events involving multiple people and conditions, express each probability in terms of unknowns and solve step-by-step using conditional and joint probability rules.

Answer 1

The Correct Option is A

Step 1: Let the individual success probabilities be \[ P(S_1) = p_1, \quad P(S_2) = p_2, \quad P(S_3) = p_3 \] Step 2: Use the formula for event \( U \) \[ P(U) = 1 - (1 - p_1)(1 - p_2)(1 - p_3) = \frac{1}{2} \Rightarrow (1 - p_1)(1 - p_2)(1 - p_3) = \frac{1}{2} \quad \cdots (1) \] Step 3: Use the conditional probability for \( V \) \[ P(V) = \frac{P(S_1 \cap S_2' \cap S_3')}{P(S_2' \cap S_3')} = \frac{p_1(1 - p_2)(1 - p_3)}{(1 - p_2)(1 - p_3)} = p_1 = \frac{1}{10} \quad \cdots (2) \] Step 4: Use the definition of \( W \) \[ P(W) = p_2(1 - p_3) = \frac{1}{12} \quad \cdots (3) \] Step 5: Let \( x = 1 - p_3 \) From (3), \[ p_2 x = \frac{1}{12} \quad \Rightarrow \quad p_2 = \frac{1}{12x} \] From (1), \[ (1 - p_1)(1 - p_2)(1 - p_3) = (1 - \frac{1}{10})(1 - \frac{1}{12x})x = \frac{9}{10}\left(1 - \frac{1}{12x}\right)x = \frac{1}{2} \] Step 6: Solve the equation \[ \frac{9}{10} \left(1 - \frac{1}{12x} \right) x = \frac{1}{2} \Rightarrow \frac{9}{10} \left( x - \frac{1}{12} \right) = \frac{1}{2} \Rightarrow \frac{9x}{10} - \frac{3}{40} = \frac{1}{2}\] \[\Rightarrow \frac{9x}{10} = \frac{1}{2} + \frac{3}{40} = \frac{23}{40} \Rightarrow x = \frac{23}{36} \Rightarrow 1 - p_3 = \frac{23}{36} \Rightarrow p_3 = \frac{13}{36} \]