3. A problem is given to three students A, B, C whose chances of solving it are 1 2 , 1 3 , 1 4 respectively. What is the probability that (i) The problem is solved. (ii) Exactly one of them solves the problem

We are given the probabilities that students A, B, and C will solve a problem. Their respective chances are:

- A: \( P(A) = \frac{1}{2} \)
- B: \( P(B) = \frac{1}{3} \)
- C: \( P(C) = \frac{1}{4} \)

We will calculate the probability that:

(i) **The problem is solved**  
(ii) **Exactly one of them solves the problem**

### (i) Probability that the problem is solved

The probability that at least one student solves the problem is the complement of the probability that none of them solves the problem.

- Probability that A **does not** solve the problem:  
  \( P(\text{A does not solve}) = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)

- Probability that B **does not** solve the problem:  
  \( P(\text{B does not solve}) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)

- Probability that C **does not** solve the problem:  
  \( P(\text{C does not solve}) = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4} \)

Thus, the probability that **none of them solves the problem** is the product of the individual probabilities that they don't solve it:

\[
P(\text{None solve}) = P(\text{A does not solve}) \times P(\text{B does not solve}) \times P(\text{C does not solve})
\]
\[
P(\text{None solve}) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}
\]

Therefore, the probability that **at least one of them solves the problem** is:

\[
P(\text{At least one solves}) = 1 - P(\text{None solve}) = 1 - \frac{1}{4} = \frac{3}{4}
\]

### (ii) Probability that exactly one of them solves the problem

For exactly one student to solve the problem, one student must solve it while the other two fail. We calculate this for each student and then sum the probabilities.

1. **A solves it, but B and C do not**:
   \[
   P(\text{A solves, B and C do not}) = P(A) \times P(\text{B does not solve}) \times P(\text{C does not solve})
   \]
   \[
   = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}
   \]

2. **B solves it, but A and C do not**:
   \[
   P(\text{B solves, A and C do not}) = P(B) \times P(\text{A does not solve}) \times P(\text{C does not solve})
   \]
   \[
   = \frac{1}{3} \times \frac{1}{2} \times \frac{3}{4} = \frac{1}{8}
   \]

3. **C solves it, but A and B do not**:
   \[
   P(\text{C solves, A and B do not}) = P(C) \times P(\text{A does not solve}) \times P(\text{B does not solve})
   \]
   \[
   = \frac{1}{4} \times \frac{1}{2} \times \frac{2}{3} = \frac{1}{12}
   \]

Finally, the probability that exactly one of them solves the problem is the sum of these three cases:

\[
P(\text{Exactly one solves}) = \frac{1}{4} + \frac{1}{8} + \frac{1}{12}
\]

Finding the common denominator (LCM of 4, 8, and 12 is 24):

\[
P(\text{Exactly one solves}) = \frac{6}{24} + \frac{3}{24} + \frac{2}{24} = \frac{11}{24}
\]

### Summary of Results:
(i) The probability that the problem is solved: \( \frac{3}{4} \)

(ii) The probability that exactly one of them solves the problem: \( \frac{11}{24} \)

Would you like more details or have any other questions?

Here are 5 related questions:
1. What is the probability that **exactly two** students solve the problem?
2. How would the solution change if A's chance of solving the problem was \( \frac{2}{3} \)?
3. How do probabilities change if the number of students is increased to four?
4. What is the probability that **none** of the students solve the problem?
5. What if the problem was repeated and you wanted the probability that at least one student solves both attempts?

**Tip:** When solving probability problems, the complement rule is often very useful to simplify calculations for "at least one" scenarios.

A problem is given to three students A, B, C whose chances of solving it are 1/2, 1/3, 1/4 respectively. What is the probability that (i) The problem is solved. (ii) Exactly one of them solves the problem.

Explore how to solve a probability problem involving three students with different chances of solving it. This problem demonstrates key concepts like independent events, complement rule, and exact probability calculations. Ideal for high school students in Grades 9-12.

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