Question

a horse race has 12 entries and one person owns 2 of those horses. assuming that there are no ties, what is the probability that those two horses finish first and second (regardless of order)?
the probability that the two horses finish first and second is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the total number of ways to choose 2 horses for the first two positions

The number of permutations of \(n = 12\) horses taken \(r=2\) at a time is given by the formula \(P(n,r)=\frac{n!}{(n - r)!}\).
So, \(P(12,2)=\frac{12!}{(12 - 2)!}=\frac{12!}{10!}=12\times11 = 132\).

Step2: Calculate the number of favorable cases

The number of ways the two - owned horses can finish first and second (regardless of order) is \(n = 2\) (since the two horses can swap positions: horse A first and horse B second or horse B first and horse A second).

Step3: Calculate the probability

The probability \(P\) is the number of favorable outcomes divided by the number of total outcomes.
\(P=\frac{2}{132}=\frac{1}{66}\approx0.0152\)

Answer:

\(0.0152\)