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

To find the probability \( P(T) \), we start by considering the definition of each event and how they interrelate:

• \( U \): 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 it.
• \( W \): \( S_2 \) can solve the problem and \( S_3 \) cannot solve the problem.
• \( T \): \( S_3 \) can solve the problem.

We have the following probabilities given:

• \( P(U) = \frac{1}{2} \)
• \( P(V) = \frac{1}{10} \)
• \( P(W) = \frac{1}{12} \)

Since \( U \) is the event that at least one of the students can solve the problem, we also have the complementary event \( \bar{U} \), which is that none can solve it.

The probability of \( U \) can be broken down as:

\( P(U) = 1 - P(\bar{U}) = 1 - \left( 1-P(T)-P(V)-P(W) \right) \)

Rearranging, we get:

\( P(U) = 1 - \left( 1 - P(T) - \frac{1}{10} - \frac{1}{12} \right) \)

We simplify the equation knowing \( P(U) = \frac{1}{2} \):

\( \frac{1}{2} = 1 - \left( 1 - P(T) - \frac{1}{10} - \frac{1}{12} \right) \)

\( \frac{1}{2} = P(T) + \frac{1}{10} + \frac{1}{12} \)

We need a common denominator to add the fractions. The least common multiple of 10 and 12 is 60:

\( \frac{1}{2} = P(T) + \frac{6}{60} + \frac{5}{60} \)

\( \frac{1}{2} = P(T) + \frac{11}{60} \)

Express \( \frac{1}{2} \) as \( \frac{30}{60} \) to have a common denominator:

\( \frac{30}{60} = P(T) + \frac{11}{60} \)

Subtract \( \frac{11}{60} \) from both sides:

\( P(T) = \frac{30}{60} - \frac{11}{60} = \frac{19}{60} \)

Therefore, the probability \( P(T) \) is: \( \frac{19}{60} \)