+0

# Places on a Podium Probaility

0
75
9

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?

Nov 27, 2020

#1
-1

10 c 3 = 120 ways to choose 3 places

but only ONE is correct order 1-2-3

1/120

Nov 27, 2020
#2
0

I just checked and the answer should have been 1/720... Why might that be?

WillyGolden  Nov 27, 2020
#4
+1

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 !

Guest Nov 27, 2020
#6
0

that should be 1/6  x 1/120 = 1/720     jeez.....

Guest Nov 27, 2020
#3
+1

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

Nov 27, 2020
#5
0

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!

WillyGolden  Nov 27, 2020
#7
0

10! / (10-3)! = 10! / 7! = 720     One of which is correct order 1 - 2 - 3           =   1/720

Guest Nov 27, 2020
#9
0

This is a permutation, as order matters.

Keep in mind that that permutation will only give you the possible, but you can do combinations when there is only one possible arrangement as it returns only one arrangment: 3 choose 3 = 1, whereas 3 pick 3 = 6

Pangolin14  Nov 27, 2020
#8
0

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.

720 possible.

For the favorable, we have 1 way to choose 1st, 1 way to choose 2nd, 1 way to choose 3rd.

So 1/720

Nov 27, 2020