Question 1045370: A horse race has 11 entries and one person owns 4 of those horses. Assuming that there are no ties, what is the probability that those four horses finish first, second, third, and fourth?(regardless of order)(Round to four decimal places as needed.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there are 11 entries in total.
the owner has 4 horses.
if each horse has the same probability of winning, then the probability that the owner's horses will come in first, second, third, and fourth will be as follows:
note that it was not specified which horse had to be first, second, third, or fourth.
it was only specified that one of those horses needed to be in each position.
this makes it a combination type problem rather than a permutation type problem because order doesn't matter.
we have 4 spots.
the probability that one of those 4 horses will be first is 4/11.
the probability that one of the remaining 3 horses will be second is 3/10.
the probability that one of the remaining 2 horses will be third is 2/9.
the probability that the remaining horse will be fourth is 1/8.
the total probability is 4/11 * 3/10 * 2/9 * 1/8 = .003030303
that's equivalent to 1/330
you can also look at is as the number of ways you can 4 out of 4 divided by the number of ways you can get 4 out of 11.
that would be c(4,4) / c(11,4) = 1 / 330.
c(n,x) equals number of ways you can get x elements out of n.
c(n,x) = n! / (x! * (n-x)!)
c(4,4) = 4! / (4! * 0!) = 4! / 4! = 1
c(11,4) = 11! / (4! * 7!) = (11 * 10 * 9 * 8 * 7!) / (4! * 7!) = (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1) = 330.
note that 0! = 1 by definition.
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