Probability of not choosing consecutive numbers from 1 to 30

To calculate the probability of not choosing consecutive numbers from 1 to 30, we need to determine the total number of possible outcomes and the number of favorable outcomes.

Total number of possible outcomes:
Since we are choosing 3 numbers from 1 to 30, the total number of possible outcomes can be calculated using the combination formula:

nCr = n! / (r!(n-r)!)

In this case, n is the total number of choices (30) and r is the number of selections (3). So, the total number of possible outcomes is:

30C3 = 30! / (3!(30-3)!) = 30! / (3!27!) = (30 * 29 * 28) / (3 * 2 * 1) = 4060

Therefore, there are 4060 possible outcomes when choosing 3 numbers from 1 to 30.

Number of favorable outcomes:
To determine the number of favorable outcomes, we need to consider the cases where the chosen numbers are not consecutive.

Case 1: No consecutive numbers from 1 to 30
In this case, we need to exclude the numbers that are adjacent to each other. For example, if we choose 1, 2, and 3, it is not a favorable outcome. Similarly, if we choose 2, 3, and 4, it is also not a favorable outcome.

There are 28 pairs of consecutive numbers from 1 to 30 (1-2, 2-3, 3-4, ..., 27-28, 28-29, 29-30). We need to exclude these pairs from the total number of possible outcomes.

Since we have 28 pairs and we are choosing 3 numbers from each pair, the number of favorable outcomes for this case is:

28C3 = 28! / (3!(28-3)!) = 28! / (3!25!) = (28 * 27 * 26) / (3 * 2 * 1) = 3276

Case 2: Consecutive numbers at the beginning or end
In this case, we need to exclude the numbers that are adjacent to the chosen numbers at the beginning or end. For example, if we choose 1, 2, and 7, it is not a favorable outcome. Similarly, if we choose 25, 29, and 30, it is also not a favorable outcome.

There are 2 pairs of consecutive numbers at the beginning and end (1-2 and 29-30). We need to exclude these pairs from the total number of possible outcomes.

Since we have 2 pairs and we are choosing 3 numbers from each pair, the number of favorable outcomes for this case is:

2C3 = 2! / (3!(2-3)!) = 2! / (3!(-1)!) = 0

Total number of favorable outcomes:
To calculate the total number of favorable outcomes, we sum the favorable outcomes from both cases:

Total favorable outcomes = favorable outcomes from Case 1 + favorable outcomes from Case 2 = 3276 + 0 = 3276

Probability:
Finally