Six people, including a, b, and c, form a queue in a random order (all 6! orderings are equiprobable). consider the event "b is between a and c in the queue". what is its probability? (the order of a and c can be arbitrary, but b should be between them)

There is a 1.39% probability that "B is between A and C in the queue".

Step-by-step explanation:

The first step to solve this problem is find the total number of possible orderings:

The first person of the queue can be any of the six. The second, can be any but the first, so five.

So

There are 6*5*4*3*2*1 = 720 total queue orderings.

Now we find the number of queues that B is between A and C. So:

We have:

B as the second, A as the first and C as the third

B as the second, C as the first and A as the third

B as the third, A as the second and C as the fourth

B as the third, C as the second and A as the fourth

B as the fourth, A as the third and C as the fifth.

B as the fourth, C as the third and A as the fifth.

B as the fifth, A as the fourth and C as the sixth.

B as the fifth, C as the fourth and A as the sixth.

So 8 total outcomes in which B is between A and C.

What is its probability?

There is a 1.11% probability that "B is between A and C in the queue".

how many players are there?

step-by-step explanation:

step-by-step explanation:

you can add the two numbers and divide the result by 2. or you can find the gap between the two numbers and add half of this to the smaller number.