Probability Of Drawing Three Even Cards From A Set Of Five

In the realm of mathematics, probability serves as a powerful tool for quantifying the likelihood of specific events occurring. It finds widespread application in various scenarios, from predicting weather patterns to assessing the odds in games of chance. Card games, in particular, provide a fertile ground for exploring probability concepts, as the random nature of card draws introduces an element of uncertainty that can be analyzed using mathematical principles. In this comprehensive guide, we will delve into a specific probability problem involving card selection, dissecting the steps involved in arriving at the solution and providing a clear understanding of the underlying concepts.

This article aims to provide a detailed explanation of how to calculate the probability of a specific event in a card game scenario. We'll break down the problem step-by-step, ensuring that readers can follow the logic and apply it to similar situations. This problem involves calculating the probability of Chelsea drawing three even-numbered cards from a set of five cards. By the end of this guide, you'll have a solid understanding of the principles of combinations and probability calculations, enabling you to tackle similar problems with confidence.

Let's set the stage for our probability exploration with a specific scenario. Imagine Chelsea is engrossed in a card game where she faces a crucial decision. She has a selection of 5 cards to choose from, and her task is to pick 3 of them. The twist? We want to determine the probability that all three cards she chooses are even-numbered. This problem introduces us to the core concepts of combinations and probability, which we will unravel step by step.

To solve this, we need to understand the fundamental concepts of probability and combinations. Probability is the measure of the likelihood that an event will occur. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. The formula for probability is:

$$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

Combinations, on the other hand, deal with the selection of items from a set where the order of selection does not matter. The number of ways to choose k items from a set of n items is given by the combination formula:

$$C(n, k) = \frac{n!}{k!(n-k)!}$$

Where "!" denotes the factorial function, which is the product of all positive integers up to that number (e.g., 5! = 5 × 4 × 3 × 2 × 1 = 120). Understanding these concepts is crucial for solving probability problems involving combinations, as it allows us to accurately calculate the number of possible outcomes and the number of favorable outcomes.

The first crucial step in tackling any probability problem is to define the sample space. The sample space represents the set of all possible outcomes of an experiment. In Chelsea's card game, the experiment is choosing 3 cards out of 5. To define our sample space, we need to list out all the possible combinations of 3 cards that can be selected from the 5 available cards. This is where the concept of combinations comes into play.

To begin, let's assume the five cards are numbered 1, 2, 3, 4, and 5. We are interested in the probability that all three cards Chelsea chooses are even numbers. Therefore, we must first identify which cards in our set are even. In this case, the even-numbered cards are 2 and 4. This is a crucial piece of information, as it sets the stage for calculating the probability of selecting only even-numbered cards. By clearly defining the even cards within our sample space, we narrow our focus to the outcomes that meet our specific criteria.

Now, we need to determine the total number of ways Chelsea can choose 3 cards from the 5 available cards, regardless of whether they are even or odd. This involves using the combination formula, which, as we discussed earlier, is:

$$C(n, k) = \frac{n!}{k!(n-k)!}$$

In our case, n represents the total number of cards (5), and k represents the number of cards Chelsea chooses (3). Plugging these values into the formula, we get:

$$C(5, 3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}$$

Let's break this down step by step: 5! (5 factorial) is 5 × 4 × 3 × 2 × 1 = 120, 3! (3 factorial) is 3 × 2 × 1 = 6, and 2! (2 factorial) is 2 × 1 = 2. Substituting these values, we have:

$$C(5, 3) = \frac{120}{6 × 2} = \frac{120}{12} = 10$$

Therefore, there are 10 possible ways to choose 3 cards from a set of 5 cards. This calculation forms the denominator of our probability fraction, representing the total number of possible outcomes in our sample space.

The next crucial step is to determine the number of ways Chelsea can choose 3 even-numbered cards. Recall that we identified two even-numbered cards in our set: 2 and 4. However, Chelsea needs to choose 3 cards, and we only have 2 even cards available. This means it is impossible for Chelsea to choose 3 even cards from this set.

Mathematically, this can be represented as choosing 3 items from a set of 2, which is C(2, 3). Using the combination formula:

$$C(2, 3) = \frac{2!}{3!(2-3)!}$$

Here, we encounter a problem: we have (2-3)! which is (-1)!. Factorials are only defined for non-negative integers, so (-1)! is undefined. Moreover, intuitively, you cannot choose 3 items from a set of 2. Therefore, the number of ways to choose 3 even cards is 0. This realization is critical for the final probability calculation.

Now that we have determined both the total number of possible outcomes and the number of favorable outcomes (choosing 3 even cards), we can calculate the probability. Recall the probability formula:

$$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

We found that the number of ways to choose 3 even cards is 0, and the total number of ways to choose 3 cards is 10. Plugging these values into the formula, we get:

$$\text{Probability} = \frac{0}{10} = 0$$

Therefore, the probability that Chelsea will choose 3 even cards is 0. This result makes intuitive sense because there are only two even-numbered cards in the set, making it impossible to choose three even cards.

In conclusion, the probability that Chelsea will choose 3 even-numbered cards from the given set of 5 cards (1, 2, 3, 4, 5) is 0. This result stems from the fact that there are only two even-numbered cards in the set, making it impossible to select three even cards. This problem highlights the importance of understanding the fundamental principles of probability and combinations, as well as carefully defining the sample space and identifying favorable outcomes. By breaking down the problem into smaller, manageable steps, we can arrive at a clear and accurate solution.

None of the provided options (a. $rac{3}{5}$, c. $rac{3}{10}$) are correct. The correct answer is 0.

Probability, card game, combinations, even cards, sample space, favorable outcomes, total outcomes, mathematics.