Three persons A, B and C speak at a function along with 5 other persons. If the persons speak at random, find the probability that A speaks before B and B speaks before C

Options

A. `3/8`
B. `1/6`
C. `3/5`
D. `5/6`

Answer

To solve the problem, we need to find the probability that person A speaks before person B, and person B speaks before person C when all three are speaking at a function along with 5 other persons.

### Step-by-Step Solution:

1. **Total Number of Speakers**:
There are 3 specific persons (A, B, C) and 5 other persons, making a total of 8 persons.

2. **Total Arrangements**:
The total number of ways to arrange 8 persons is given by the factorial of the number of persons, which is:
\[ 8! = 40320 \]

3. **Favorable Outcomes**:
We want to find the arrangements where A speaks before B, and B speaks before C.
- The three persons A, B, and C can be arranged in any order among themselves. The total arrangements of A, B, and C is \(3!\) (which is 6).
- However, we are only interested in the arrangement where A is before B and B is before C. There is only **1** favorable arrangement (A, B, C).

4. **Arranging Other Persons**:
The remaining 5 persons can be arranged in any order. The number of arrangements for these 5 persons is:
\[ 5! = 120 \]

5. **Total Favorable Outcomes**:
The total number of favorable outcomes where A is before B and B is before C is:
\[ 5! \times 1 = 120 \]

6. **Probability Calculation**:
The probability that A speaks before B and B speaks before C is given by the ratio of favorable outcomes to total outcomes:
\[ P(A \text{ before } B \text{ and } B \text{ before } C) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{120}{40320} \]
Simplifying this gives:
\[ P = \frac{1}{336} \]

### Final Answer:
The probability that A speaks before B and B speaks before C is:
\[ \frac{1}{6} \]