Final answer:
The probability of randomly selecting three horses that finish in the top three positions in any order in a field of 10 horses is 1 in 120, or 0.0083 when rounded to four decimal places.
Step-by-step explanation:
In a horse race, when betting on the three horses that finish first, second, and third in any order, you're making a type of wager known as a 'trifecta box'. In this scenario, given a field of 10 horses, each horse has an equal chance of finishing in any of the top three positions. To calculate the probability of winning this bet with a random selection of three horses, we can use the principles of combinatorics.
First, we determine how many different combinations of three horses can come in the top three positions out of 10 horses. This is a combination problem (since the order does not matter), calculated using the formula for combinations: C(n, k) = n! / [k! * (n - k)!], where n is the total number of items, k is the number of items to choose, and '!' denotes factorial.
Using this formula, the number of combinations for the top three positions from 10 horses is C(10, 3) = 10! / [3! * (7!)] = 120. Therefore, there are 120 different ways the three horses could finish in the top three spots.
The probability of selecting the correct combination of three horses, therefore, is 1 out of these 120 combinations. Expressed as a decimal, this is a probability of 0.0083 (rounded to four decimal places).