I am trying to answer the following question:
"What is the probability that you guess (at random, in correct order) the 1st, 2nd and 3rd place winners before a race with 10 contestants?"
My guess is that this is a permutation problem since order matters, but I'm unsure of how to use the formula in this case. Would I multiply the outcome of three different possible permutations?
10 c 3 = 120 ways to choose 3 places
but only ONE is correct order 1-2-3
I just checked and the answer should have been 1/720... Why might that be?
I see my error
10 c 3 can be 1 2 3 3 2 1 2 1 3 321 312 1 32
so only 1 of the choices would be correct
1/6 x 120 = 1/720
or 10 P 3 = 720
Sorry for the confusion !
Suppose all ten are lined up randomly in front of you and you have to pin the winning ribbons on the first three - BUT, you've forgotten who they are!
What is the probability that you will just happen to pin the winners ribbon on the overall winner? 1/10
Then, what's the probability that, out of the remaining nine, you will just happen to pin the second place ribbon on the right person? 1/9
Then what's the probability that, out of the remaining eight, you will just happen to pin the third place ribbon on the right person? 1/8
So overall your probability of getting all three right is (1/10)*(1/9)*(1/8) = 1/720
Thank you for your explanation! So was this an example of a combination or of permutaion? I tried both formulas:
Combination = n! / (n - r)! r!
Permutation = n! / (n - r)!
But neither gave the answer you found. I was using:
n = 10
r = 3
Is that correct? If so, was I just calculating incorrectly? Doesn't 10! = 10 x 9 x 8 x 7 x 6 ... x 1?
Thank you for your help so far! I am trying to learn more about the math behind statistics and have been havign some trouble!
10! / (10-3)! = 10! / 7! = 720 One of which is correct order 1 - 2 - 3 = 1/720
Favorable / possible
Possible would be 10 contestants, then choose 3 of them.
There are 10 ways to choose first, 9 ways to choose second, 8 ways to choose 3rd.
For the favorable, we have 1 way to choose 1st, 1 way to choose 2nd, 1 way to choose 3rd.