Probability that A will pass the exam is `(1)/(4)`, B will pass the exam is `(2)/(5)` and C will pass the exam is `(2)/(3)`. The probability that exactly one of them will pass th exam is

Options

A. `(2)/(5)`
B. `(3)/(20)`
C. `(9)/(20)`
D. `(4)/(5)`

Answer

To solve the problem of finding the probability that exactly one of A, B, or C will pass the exam, we will follow these steps:

### Step 1: Define the probabilities
Let:
- \( P(A) = \frac{1}{4} \) (Probability that A will pass)
- \( P(B) = \frac{2}{5} \) (Probability that B will pass)
- \( P(C) = \frac{2}{3} \) (Probability that C will pass)

### Step 2: Calculate the probabilities of not passing
Now, we calculate the probabilities that each of them will not pass:
- \( P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4} \)
- \( P(B') = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5} \)
- \( P(C') = 1 - P(C) = 1 - \frac{2}{3} = \frac{1}{3} \)

### Step 3: Calculate the probability of exactly one passing
The probability that exactly one of them passes can be calculated by considering three cases:
1. Only A passes
2. Only B passes
3. Only C passes

We will denote the required probability as \( P \).

#### Case 1: Only A passes
The probability that only A passes is given by:
\[ P(\text{Only A passes}) = P(A) \cdot P(B') \cdot P(C') \]
\[ = \frac{1}{4} \cdot \frac{3}{5} \cdot \frac{1}{3} \]
\[ = \frac{1 \cdot 3 \cdot 1}{4 \cdot 5 \cdot 3} = \frac{3}{60} = \frac{1}{20} \]

#### Case 2: Only B passes
The probability that only B passes is given by:
\[ P(\text{Only B passes}) = P(B) \cdot P(A') \cdot P(C') \]
\[ = \frac{2}{5} \cdot \frac{3}{4} \cdot \frac{1}{3} \]
\[ = \frac{2 \cdot 3 \cdot 1}{5 \cdot 4 \cdot 3} = \frac{6}{60} = \frac{1}{10} = \frac{2}{20} \]

#### Case 3: Only C passes
The probability that only C passes is given by:
\[ P(\text{Only C passes}) = P(C) \cdot P(A') \cdot P(B') \]
\[ = \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{3}{5} \]
\[ = \frac{2 \cdot 3 \cdot 3}{3 \cdot 4 \cdot 5} = \frac{18}{60} = \frac{3}{10} = \frac{6}{20} \]

### Step 4: Sum the probabilities
Now, we sum the probabilities of all three cases:
\[ P = P(\text{Only A passes}) + P(\text{Only B passes}) + P(\text{Only C passes}) \]
\[ = \frac{1}{20} + \frac{2}{20} + \frac{6}{20} \]
\[ = \frac{1 + 2 + 6}{20} = \frac{9}{20} \]

### Final Result
Thus, the probability that exactly one of them will pass the exam is:
\[ \boxed{\frac{9}{20}} \]