Three numbers are chosen at random from 1 to 15. The probability that they are consecutive is

Options

A. `1/190`
B. `1/120`
C. `1/35`
D. `5/75`

Answer

To find the probability that three numbers chosen at random from the set of numbers 1 to 15 are consecutive, we can follow these steps:

### Step 1: Determine the total number of ways to choose 3 numbers from 15.
The total number of ways to select 3 numbers from a set of 15 can be calculated using the combination formula:

\[ \text{Total ways} = \binom{15}{3} = \frac{15!}{3!(15-3)!} \]

Calculating this gives:

\[ \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455 \]

### Step 2: Determine the number of favorable outcomes (consecutive numbers).
Consecutive numbers can be represented as (n, n+1, n+2). The smallest value for n is 1 (giving us 1, 2, 3) and the largest value for n is 13 (giving us 13, 14, 15).

Thus, the possible sets of consecutive numbers are:
- (1, 2, 3)
- (2, 3, 4)
- (3, 4, 5)
- (4, 5, 6)
- (5, 6, 7)
- (6, 7, 8)
- (7, 8, 9)
- (8, 9, 10)
- (9, 10, 11)
- (10, 11, 12)
- (11, 12, 13)
- (12, 13, 14)
- (13, 14, 15)

Counting these, we find there are 13 sets of consecutive numbers.

### Step 3: Calculate the probability.
The probability of selecting 3 consecutive numbers is given by the ratio of the number of favorable outcomes to the total number of outcomes:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{13}{455} \]

### Step 4: Simplify the probability.
To simplify \(\frac{13}{455}\), we can check if there are any common factors. Since 13 is a prime number and does not divide 455, the fraction is already in its simplest form.

Thus, the final answer for the probability that three randomly chosen numbers from 1 to 15 are consecutive is:

\[ \frac{1}{35} \]