A problem in Mathematics is given to two students A, B and their chances of solving the problem are `2/3, 3/4` respectively. The probability that exactly one of them solves the problem is

Options

A. `1/2`
B. `5/6`
C. `7/12`
D. `5/12`

Answer

To find the probability that exactly one of the students A or B solves the problem, we can follow these steps:

### Step 1: Identify the probabilities
Given:
- Probability that student A solves the problem, \( P(A) = \frac{2}{3} \)
- Probability that student B solves the problem, \( P(B) = \frac{3}{4} \)

### Step 2: Calculate the probabilities of not solving the problem
- Probability that student A does not solve the problem, \( P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3} \)
- Probability that student B does not solve the problem, \( P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \)

### Step 3: Define the scenarios for exactly one solving the problem
There are two scenarios where exactly one of them solves the problem:
1. A solves the problem and B does not.
2. B solves the problem and A does not.

### Step 4: Calculate the probabilities for each scenario
1. Probability that A solves and B does not:
\[ P(A \text{ and } B') = P(A) \times P(B') = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6} \]

2. Probability that B solves and A does not:
\[ P(B \text{ and } A') = P(B) \times P(A') = \frac{3}{4} \times \frac{1}{3} = \frac{3}{12} = \frac{1}{4} \]

### Step 5: Add the probabilities of the two scenarios
To find the total probability that exactly one of them solves the problem, we add the probabilities from both scenarios:
\[ P(\text{exactly one solves}) = P(A \text{ and } B') + P(B \text{ and } A') = \frac{1}{6} + \frac{1}{4} \]

### Step 6: Find a common denominator and combine
The common denominator for 6 and 4 is 12:
\[ P(\text{exactly one solves}) = \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \]

### Final Answer
The probability that exactly one of them solves the problem is \( \frac{5}{12} \).

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