To find the probability that exactly two of A, B, and C solve the problem, we will follow these steps:
### Step 1: Identify the probabilities
The probabilities of A, B, and C solving the problem are given as follows:
- Probability of A solving the problem, \( P(A) = \frac{1}{2} \)
- Probability of B solving the problem, \( P(B) = \frac{1}{3} \)
- Probability of C solving the problem, \( P(C) = \frac{1}{4} \)
### Step 2: Calculate the probabilities of not solving the problem
Next, we need to calculate the probabilities of each person not solving the problem:
- Probability of A not solving the problem, \( P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \)
- Probability of B not solving the problem, \( P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \)
- Probability of C not solving the problem, \( P(C') = 1 - P(C) = 1 - \frac{1}{4} = \frac{3}{4} \)
### Step 3: Calculate the probability for each combination where exactly two solve the problem
We need to consider three scenarios where exactly two out of the three solve the problem:
1. A and B solve the problem, but C does not:
\[
P(A \text{ and } B \text{ and } C') = P(A) \cdot P(B) \cdot P(C') = \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} = \frac{1}{8}
\]
2. A and C solve the problem, but B does not:
\[
P(A \text{ and } C \text{ and } B') = P(A) \cdot P(C) \cdot P(B') = \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{12}
\]
3. B and C solve the problem, but A does not:
\[
P(B \text{ and } C \text{ and } A') = P(B) \cdot P(C) \cdot P(A') = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{24}
\]
### Step 4: Sum the probabilities of the three scenarios
Now, we add the probabilities of these three scenarios to find the total probability that exactly two of them solve the problem:
\[
P(\text{exactly 2 solve}) = P(A \text{ and } B \text{ and } C') + P(A \text{ and } C \text{ and } B') + P(B \text{ and } C \text{ and } A'
\]
Substituting the values calculated:
\[
P(\text{exactly 2 solve}) = \frac{1}{8} + \frac{1}{12} + \frac{1}{24}
\]
### Step 5: Find a common denominator and sum the fractions
The least common multiple of 8, 12, and 24 is 24. We convert each fraction:
- \( \frac{1}{8} = \frac{3}{24} \)
- \( \frac{1}{12} = \frac{2}{24} \)
- \( \frac{1}{24} = \frac{1}{24} \)
Now, summing these:
\[
P(\text{exactly 2 solve}) = \frac{3}{24} + \frac{2}{24} + \frac{1}{24} = \frac{6}{24} = \frac{1}{4}
\]
### Final Answer
The probability that exactly two of them solve the problem is \( \frac{1}{4} \).
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