Problem 152
Question
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is [2003] (A) \(\frac{4}{5}\) (B) \(\frac{3}{5}\) (C) \(\frac{1}{5}\) (D) \(\frac{2}{5}\)
Step-by-Step Solution
Verified Answer
The probability that Mr. A selected the winning horse is \( \frac{2}{5} \).
1Step 1: Understanding the problem
Mr. A selects two horses at random from a total of five horses in a race. We need to determine the probability of him choosing at least one winning horse.
2Step 2: Calculate total number of ways to select horses
To find out the total number of ways to select 2 horses, we use the combination formula \( \binom{n}{r} \), which is \( \binom{5}{2} \). This results in \( \frac{5!}{2!(5-2)!} = 10 \) different combinations.
3Step 3: Calculate favorable outcomes
The winning probability is considering the situation where one of the horses selected is the winning horse. Since there is one winning horse, Mr. A can choose this winning horse and any of the other 4 horses. Therefore, favorable outcomes are the 4 combinations consisting of the winning horse combined with one of the other horses. Hence, there are 4 favorable outcomes.
4Step 4: Compute probability
Now, we compute the probability of Mr. A choosing the winning horse as \( \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{4}{10} = \frac{2}{5} \).
Key Concepts
CombinatoricsPermutations and CombinationsEvents and Outcomes
Combinatorics
Combinatorics is a field of mathematics that helps us count and arrange items in a group. This field is essential when dealing with problems that involve probability because it provides a structured way to find all possible combinations and arrangements of items.
In our horse-racing problem, combinatorics was used to determine both the total number of ways to choose two horses from a group of five, as well as the favorable ways to include the winning horse in the selection.
Methods involved in combinatorics include permutations and combinations. While permutations consider the order of arrangement, combinations focus solely on the selection of items, regardless of order. This makes combinations crucial in situations like our problem, where the arrangement of the two horses doesn't matter.
In our horse-racing problem, combinatorics was used to determine both the total number of ways to choose two horses from a group of five, as well as the favorable ways to include the winning horse in the selection.
Methods involved in combinatorics include permutations and combinations. While permutations consider the order of arrangement, combinations focus solely on the selection of items, regardless of order. This makes combinations crucial in situations like our problem, where the arrangement of the two horses doesn't matter.
Permutations and Combinations
Permutations and combinations are critical concepts within combinatorics. They allow us to calculate the different ways to arrange or select a group of items. Understanding these is pivotal for solving probability problems and other mathematical problems involving arrangement and selection.
**Permutations**: This refers to the arrangement of items where order matters. For example, arranging 5 books on a shelf is a permutation problem because the order of books affects the arrangement.
**Combinations**: Conversely, combinations refer to selecting items where order does not matter. In the horse racing problem, Mr. A's selection of two horses is a combination because it doesn't matter which horse he bets on first or second. The combination formula, used to calculate the number of ways to select 2 horses out of 5, is given by \( \binom{n}{r} \). For this exercise, \( \binom{5}{2} = 10 \). This tells us there are 10 possible ways Mr. A could choose 2 horses from 5.
**Permutations**: This refers to the arrangement of items where order matters. For example, arranging 5 books on a shelf is a permutation problem because the order of books affects the arrangement.
**Combinations**: Conversely, combinations refer to selecting items where order does not matter. In the horse racing problem, Mr. A's selection of two horses is a combination because it doesn't matter which horse he bets on first or second. The combination formula, used to calculate the number of ways to select 2 horses out of 5, is given by \( \binom{n}{r} \). For this exercise, \( \binom{5}{2} = 10 \). This tells us there are 10 possible ways Mr. A could choose 2 horses from 5.
Events and Outcomes
When dealing with probability, it is important to distinguish between events and outcomes. These concepts help us identify what constitutes a favorable result in a probability problem.
**Outcomes**: These are the possible results that can occur when an experiment is performed. In our example, an outcome refers to any specific pair of horses Mr. A selects. There are 10 possible outcomes when choosing 2 horses from 5.
**Events**: An event is a specific set of favorable outcomes that meets the criteria we are interested in. For Mr. A, a favorable event is selecting a pair that includes the winning horse. Only 4 of the original 10 outcomes consist of such an event, moving us to the calculation of Mr. A's probability, which is the ratio of favorable outcomes to the total number of outcomes: \( \frac{4}{10} = \frac{2}{5} \).
Understanding these concepts is essential when calculating probabilities, as it helps us identify and calculate both the total and favorable outcomes clearly and accurately.
**Outcomes**: These are the possible results that can occur when an experiment is performed. In our example, an outcome refers to any specific pair of horses Mr. A selects. There are 10 possible outcomes when choosing 2 horses from 5.
**Events**: An event is a specific set of favorable outcomes that meets the criteria we are interested in. For Mr. A, a favorable event is selecting a pair that includes the winning horse. Only 4 of the original 10 outcomes consist of such an event, moving us to the calculation of Mr. A's probability, which is the ratio of favorable outcomes to the total number of outcomes: \( \frac{4}{10} = \frac{2}{5} \).
Understanding these concepts is essential when calculating probabilities, as it helps us identify and calculate both the total and favorable outcomes clearly and accurately.
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