What is the probability of accurately picking the winner and runner-up from a race of 4 horses, where the order matters?

Study for the SAE Mathematics Exam. Study with flashcards and multiple choice questions, each question has hints and explanations. Get ready for your exam!

Multiple Choice

What is the probability of accurately picking the winner and runner-up from a race of 4 horses, where the order matters?

Explanation:
To determine the probability of accurately picking both the winner and the runner-up from a race involving 4 horses, we must consider the total possible outcomes and the specific favorable outcomes. In this scenario, we want to find the number of ways to choose and arrange the winner and the runner-up from the 4 horses. The order matters because winning first is different from finishing second. To calculate this, we can choose the winner in 4 different ways (any one of the 4 horses), and then once the winner has been selected, there are 3 remaining horses to choose from for the runner-up position. Thus, the total number of favorable outcomes (ways to choose both a winner and a runner-up) is: 4 (choices for the winner) × 3 (choices for the runner-up) = 12. Next, to find the total number of possible outcomes when selecting 2 horses from the 4, we must consider that every arrangement of the horses matters, so we treat this as a permutation problem. The total number of ways to select 2 horses from 4 (considering order) can also be calculated using the permutation formula \( P(n, r) = \frac{n!}{(n-r)!} \

To determine the probability of accurately picking both the winner and the runner-up from a race involving 4 horses, we must consider the total possible outcomes and the specific favorable outcomes.

In this scenario, we want to find the number of ways to choose and arrange the winner and the runner-up from the 4 horses. The order matters because winning first is different from finishing second.

To calculate this, we can choose the winner in 4 different ways (any one of the 4 horses), and then once the winner has been selected, there are 3 remaining horses to choose from for the runner-up position. Thus, the total number of favorable outcomes (ways to choose both a winner and a runner-up) is:

4 (choices for the winner) × 3 (choices for the runner-up) = 12.

Next, to find the total number of possible outcomes when selecting 2 horses from the 4, we must consider that every arrangement of the horses matters, so we treat this as a permutation problem. The total number of ways to select 2 horses from 4 (considering order) can also be calculated using the permutation formula ( P(n, r) = \frac{n!}{(n-r)!} \

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