Which two FRACTIONS below have a DIFFERENCE of 27/40?
2 answers:
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Answer:
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".
I believe this is how you do it
