Three mountaineers Amit ,Vinit and Nishit are climbing up a mountain with their respective probabilities of reaching the summit being `(1)/(3),(1)/(5)and(1)/(4)` respectively .What is the probability that Exactly one of them reaches Summit ?

Options

A. A)`(13)/(30)`
B. B)`(17)/(30)`
C. C)`(19)/(30)`
D. D)`(11)/(30)`

Answer

To solve the problem of finding the probability that exactly one of the three mountaineers (Amit, Vinit, and Nishit) reaches the summit, we will follow these steps:

### Step 1: Identify the probabilities
- The probability that Amit reaches the summit, \( P(A) = \frac{1}{3} \)
- The probability that Vinit reaches the summit, \( P(V) = \frac{1}{5} \)
- The probability that Nishit reaches the summit, \( P(N) = \frac{1}{4} \)

### Step 2: Calculate the probabilities of not reaching the summit
- The probability that Amit does not reach the summit, \( P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \)
- The probability that Vinit does not reach the summit, \( P(V') = 1 - P(V) = 1 - \frac{1}{5} = \frac{4}{5} \)
- The probability that Nishit does not reach the summit, \( P(N') = 1 - P(N) = 1 - \frac{1}{4} = \frac{3}{4} \)

### Step 3: Calculate the probability for each scenario where exactly one mountaineer reaches the summit
1. **Amit reaches, Vinit and Nishit do not reach:**
\[ P(A \cap V' \cap N') = P(A) \times P(V') \times P(N') = \frac{1}{3} \times \frac{4}{5} \times \frac{3}{4} \]
Simplifying this:
\[ = \frac{1}{3} \times \frac{4}{5} \times \frac{3}{4} = \frac{1 \times 4 \times 3}{3 \times 5 \times 4} = \frac{12}{60} = \frac{1}{5} \]

2. **Vinit reaches, Amit and Nishit do not reach:**
\[ P(A' \cap V \cap N') = P(A') \times P(V) \times P(N') = \frac{2}{3} \times \frac{1}{5} \times \frac{3}{4} \]
Simplifying this:
\[ = \frac{2}{3} \times \frac{1}{5} \times \frac{3}{4} = \frac{2 \times 1 \times 3}{3 \times 5 \times 4} = \frac{6}{60} = \frac{1}{10} \]

3. **Nishit reaches, Amit and Vinit do not reach:**
\[ P(A' \cap V' \cap N) = P(A') \times P(V') \times P(N) = \frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} \]
Simplifying this:
\[ = \frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} = \frac{2 \times 4 \times 1}{3 \times 5 \times 4} = \frac{8}{60} = \frac{2}{15} \]

### Step 4: Add the probabilities of all three scenarios
Now, we sum the probabilities from the three scenarios:
\[ P(\text{exactly one reaches}) = P(A \cap V' \cap N') + P(A' \cap V \cap N') + P(A' \cap V' \cap N) \]
\[ = \frac{1}{5} + \frac{1}{10} + \frac{2}{15} \]

### Step 5: Find a common denominator and sum the fractions
The least common multiple of 5, 10, and 15 is 30. We convert each fraction:
- \( \frac{1}{5} = \frac{6}{30} \)
- \( \frac{1}{10} = \frac{3}{30} \)
- \( \frac{2}{15} = \frac{4}{30} \)

Now we can sum them:
\[ = \frac{6}{30} + \frac{3}{30} + \frac{4}{30} = \frac{6 + 3 + 4}{30} = \frac{13}{30} \]

### Final Answer
The probability that exactly one of them reaches the summit is \( \frac{13}{30} \).

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