In this problem, we are asked to find the probability that only one of the two candidates, Ram and Ramesh, will be selected for a job interview. We are given their individual probabilities of selection.

Probability of Only One Selection in an Interview

To solve this, we need to consider two distinct situations where only one person gets selected, and then sum their probabilities. First, let's list the given probabilities and calculate the probabilities of non-selection.

Ram and Ramesh Probabilities

We are provided with the following probabilities for Ram's selection and Ramesh's selection:

• Probability of Ram's selection, \( P(R) = \frac{1}{6} \).
• Probability of Ramesh's selection, \( P(M) = \frac{1}{8} \).

Now, we need to find the probability that each candidate is not selected. The probability of an event not occurring is \(1\) minus the probability of the event occurring.

• Probability of Ram not being selected, \( P(R') = 1 - P(R) = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \).
• Probability of Ramesh not being selected, \( P(M') = 1 - P(M) = 1 - \frac{1}{8} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8} \).

Selection Scenarios for One Candidate

The problem asks for the probability that only one of them will be selected. This can happen in two mutually exclusive ways:

1. Ram is selected, AND Ramesh is NOT selected.
2. Ramesh is selected, AND Ram is NOT selected.

We will calculate the probability for each of these selection scenarios.

Scenario 1: Ram is Selected and Ramesh is Not Selected

Since the selection events are independent, we multiply their individual probabilities:

\( P(\text{Ram selected and Ramesh not selected}) = P(R) \times P(M') \)

\( = \frac{1}{6} \times \frac{7}{8} \)

\( = \frac{1 \times 7}{6 \times 8} \)

\( = \frac{7}{48} \)

Scenario 2: Ramesh is Selected and Ram is Not Selected

Similarly, for this scenario, we multiply their individual probabilities:

\( P(\text{Ramesh selected and Ram not selected}) = P(M) \times P(R') \)

\( = \frac{1}{8} \times \frac{5}{6} \)

\( = \frac{1 \times 5}{8 \times 6} \)

\( = \frac{5}{48} \)

One Person Probability Calculation

To find the total probability that only one of them will be selected, we add the probabilities of these two mutually exclusive scenarios:

\( P(\text{Only one selected}) = P(\text{Ram selected and Ramesh not selected}) + P(\text{Ramesh selected and Ram not selected}) \)

\( = \frac{7}{48} + \frac{5}{48} \)

\( = \frac{7 + 5}{48} \)

\( = \frac{12}{48} \)

Finally, we simplify the fraction:

\( = \frac{12 \div 12}{48 \div 12} \)

\( = \frac{1}{4} \)

Thus, the probability that only one of Ram or Ramesh will be selected is \( \frac{1}{4} \).