Horse race has 14 entrees with no ties what's probability of 4 horses owned by one owner to come in 1st, 2nd, 3rd, 4th.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
1/14 * 1/13 * 1/12 * 1/11 = ?
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To find the probability of 4 horses owned by one owner coming in 1st, 2nd, 3rd, and 4th in a horse race with 14 entrants and no ties, we need to calculate the number of favorable outcomes (where the owner's horses come in the desired positions) and divide it by the total number of possible outcomes.
First, let's calculate the number of favorable outcomes. Since the owner needs their horses to come in the 1st, 2nd, 3rd, and 4th positions, we need to consider the permutation of 4 horses out of 14. The permutation formula is:
P(n, r) = n! / (n-r)!
where n is the total number of objects and r is the number of objects taken at a time.
In this case, we have n = 14 (total horses) and r = 4 (desired positions). So, the number of favorable outcomes is:
P(14, 4) = 14! / (14-4)! = 14! / 10! = 14 * 13 * 12 * 11 = 24,024
Next, let's calculate the total number of possible outcomes. Since there are no ties in the race and 14 horses competing, we need to consider the permutation of all 14 horses. So, the total number of possible outcomes is:
P(14, 14) = 14! / (14-14)! = 14! / 0! = 14!
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 24,024 / 14!
≈ 0.000305
Therefore, the probability of 4 horses owned by one owner coming in 1st, 2nd, 3rd, and 4th in a horse race with 14 entrants and no ties is approximately 0.000305 or 0.0305%.