Out of 7 tickets consecutively numbered 3 are drawn at random. The probability for the numbers on the tickets to be in A.P. is

Options

A. 1) `(9)/(35)`
B. 2) `(""^(5)C_(3))/(""^(7)C_(3))`
C. 3) `(10)/(35)`
D. 4) `(2)/(5)`

Answer

To solve the problem, we need to find the probability that three randomly drawn tickets from a set of 7 consecutively numbered tickets are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Identify the Total Number of Tickets**:
The total number of tickets is given as 7, which are consecutively numbered (1, 2, 3, 4, 5, 6, 7).

2. **Calculate the Total Ways to Choose 3 Tickets**:
The total number of ways to choose 3 tickets from 7 is given by the combination formula:
\[ \text{Total outcomes} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

3. **Determine the Conditions for A.P.**:
For three numbers \(a\), \(b\), and \(c\) to be in A.P., the condition is:
\[ 2b = a + c \]
This implies that if we know any two numbers, the third number is determined.

4. **Count the Favorable Outcomes**:
We need to find how many combinations of 3 tickets can form an A.P.:
- The possible sets of three tickets that can form an A.P. from the numbers 1 to 7 are:
- (1, 2, 3)
- (1, 3, 5)
- (1, 4, 7)
- (2, 3, 4)
- (2, 4, 6)
- (3, 4, 5)
- (3, 5, 7)
- (4, 5, 6)
- (5, 6, 7)
- Counting these, we find that there are 9 combinations that satisfy the A.P. condition.

5. **Calculate the Probability**:
The probability \(P\) that the three drawn tickets are in A.P. is given by the ratio of the number of favorable outcomes to the total outcomes:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{9}{35} \]

### Final Answer:
The probability that the numbers on the tickets are in A.P. is \(\frac{9}{35}\).