Question

A horse race has 10 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)? The probability that the four horses finish first ,second ,third ,and fourth is …………

Answer 1

there are 10 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 4/10.
the probability that one of the remaining 3 horses will be second is 3/9.
the probability that one of the remaining 2 horses will be third is 2/8.
the probability that the remaining horse will be fourth is 1/7.

the total probability is 4/10 * 3/9 * 2/8 * 1/7 = 0.00476190
that's equivalent to 1/210