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 find 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 passes)
- \( P(B) = \frac{2}{5} \) (Probability that B passes)
- \( P(C) = \frac{2}{3} \) (Probability that C passes)

### Step 2: Calculate the probabilities of failing
The probabilities that each person fails the exam are:
- \( P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4} \)
- \( P(B') = 1 - P(B) = 1 - \frac{2}{5} = \frac{3}{5} \)
- \( P(C') = 1 - P(C) = 1 - \frac{2}{3} = \frac{1}{3} \)

### Step 3: Calculate the probability for each scenario where exactly one passes
We need to consider three scenarios where exactly one of A, B, or C passes the exam:
1. A passes, B fails, C fails: \( P(A) \cdot P(B') \cdot P(C') \)
2. A fails, B passes, C fails: \( P(A') \cdot P(B) \cdot P(C') \)
3. A fails, B fails, C passes: \( P(A') \cdot P(B') \cdot P(C) \)

Calculating each of these:
1. \( P(A) \cdot P(B') \cdot P(C') = \frac{1}{4} \cdot \frac{3}{5} \cdot \frac{1}{3} = \frac{1 \cdot 3 \cdot 1}{4 \cdot 5 \cdot 3} = \frac{3}{60} = \frac{1}{20} \)
2. \( P(A') \cdot P(B) \cdot P(C') = \frac{3}{4} \cdot \frac{2}{5} \cdot \frac{1}{3} = \frac{3 \cdot 2 \cdot 1}{4 \cdot 5 \cdot 3} = \frac{6}{60} = \frac{1}{10} \)
3. \( P(A') \cdot P(B') \cdot P(C) = \frac{3}{4} \cdot \frac{3}{5} \cdot \frac{2}{3} = \frac{3 \cdot 3 \cdot 2}{4 \cdot 5 \cdot 3} = \frac{18}{60} = \frac{3}{10} \)

### Step 4: Sum the probabilities of the three scenarios
Now, we sum the probabilities of these three scenarios to find the total probability that exactly one of them passes:
\[ P(\text{exactly one passes}) = \frac{1}{20} + \frac{1}{10} + \frac{3}{10} \]

To add these fractions, we need a common denominator. The least common multiple of 20, 10, and 10 is 20.
- Convert \( \frac{1}{10} \) to \( \frac{2}{20} \)
- Convert \( \frac{3}{10} \) to \( \frac{6}{20} \)

Now we can add:
\[ P(\text{exactly one passes}) = \frac{1}{20} + \frac{2}{20} + \frac{6}{20} = \frac{1 + 2 + 6}{20} = \frac{9}{20} \]

### Final Answer
The probability that exactly one of A, B, or C will pass the exam is \( \frac{9}{20} \).

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