Three different numbers are chosen at random from the set {1,2,3,4,5}. The numbers are arranged in increasing order. What is the probability that the resulting sequence is an arithmetic sequence?

Well, if the sequence is an arithmetic sequence, that means the difference between consecutive numbers is the same. Let's see how many possibilities we've got.

The possible increasing sequences are: (1,2,3), (1,2,4), (1,2,5), (1,3,4), (1,3,5), (1,4,5), (2,3,4), (2,3,5), (2,4,5), (3,4,5).

Out of these 10 sequences, only (1,2,3), (1,3,5), and (2,4,5) have a common difference.

Therefore, the probability of choosing an arithmetic sequence is 3/10.

Why did the arithmetic sequence ask the number sequence out? Because it needed more "common" interests!

To find the probability that the resulting sequence is an arithmetic sequence, we need to count the number of favorable outcomes (sequences that form an arithmetic sequence) and divide it by the total number of possible outcomes.

Total number of outcomes:
We are choosing 3 numbers at random from the set {1, 2, 3, 4, 5}. The total number of ways to choose 3 numbers from a set of 5 is given by the combination formula:
C(n, r) = n! / (r!(n-r)!)
where n is the total number of elements and r is the number of elements we want to choose.

In this case, n = 5 and r = 3, so the total number of outcomes is:
C(5, 3) = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 * 4 * 3!) / (3! * 2) = 10.

Favorable outcomes:
For the resulting sequence to be an arithmetic sequence, the three chosen numbers need to be consecutive. There are 3 possible ways this can occur with the given set:
1, 2, 3
2, 3, 4
3, 4, 5

So, the number of favorable outcomes is 3.

Probability:
The probability of an event happening is given by the formula:
Probability = (Number of favorable outcomes) / (Total number of outcomes)

In this case, the probability that the resulting sequence is an arithmetic sequence is:
Probability = 3 / 10 = 0.3 or 30%.

Therefore, the probability that the resulting sequence is an arithmetic sequence is 30%.

To find the probability that the resulting sequence is an arithmetic sequence, we need to determine the total number of possible outcomes and the number of favorable outcomes.

First, let's calculate the total number of possible outcomes. We have a set of 5 numbers, and we need to choose 3 different numbers without regard to their order. This can be calculated using combinations. The number of ways to choose 3 numbers from a set of 5 is denoted as "5 choose 3" and can be calculated as:

C(5, 3) = 5! / (3! * (5-3)!) = 10

Therefore, there are 10 different combinations of 3 numbers that can be chosen from the set {1, 2, 3, 4, 5}.

Now let's determine the number of favorable outcomes, i.e., the number of combinations that will result in an arithmetic sequence. For an arithmetic sequence, the numbers must have a common difference. To identify the number of arithmetic sequences, we can pick any three numbers, sort them in increasing order, and check if the differences between consecutive numbers are the same.

There are two cases to consider:

Case 1: The common difference is 1.
In this case, we can choose any three consecutive numbers from the set {1, 2, 3, 4, 5}. There are three possible combinations: {1, 2, 3}, {2, 3, 4}, and {3, 4, 5}.

Case 2: The common difference is 2.
To have a common difference of 2, we need to choose a number and then skip one number in the set {1, 2, 3, 4, 5}. There are two possibilities: {1, 3, 5} and {2, 4, 5}.

Therefore, there are a total of 5 favorable outcomes (3 from case 1 and 2 from case 2).

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = (Number of favorable outcomes) / (Number of possible outcomes) = 5 / 10 = 1/2 = 0.5

Hence, the probability that the resulting sequence is an arithmetic sequence is 0.5 or 50%.

The set is there to indicate that the three numbers chosen are distinct.

Altogether there are 6 ways to arrange 3 distinct numbers (=3!), out of which an arithmetic sequence is either
a,b,c
or
c,b,a
assuming a<b<c.
Can you calculate the probability?

An arithmetic sequence is one in which the differences between neighboring numbers is the same. The only arithmetic sequences you can get from that set of 5 numbers are: (1, 2, 3), (2, 3, 4), (3, 4, 5), or (1, 3, 5). So only 4 choices.

How many ways can you arrange 3 numbers from the set so that they are in increasing order?

123, 124, 125,
134, 135,
145,
234, 235,
245,
345.

You should be able to find the probability. Can you argue why you don't have to worry about choices like 524 and so forth?