Conditional Probability Selecting A Second Grader Explained

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Introduction

In the realm of probability, conditional probability plays a pivotal role in understanding the likelihood of an event occurring given that another event has already transpired. This concept is particularly useful in scenarios where events are not independent, and the occurrence of one event influences the probability of another. This article delves into a practical example of conditional probability, focusing on the scenario of selecting children from different grades in a room. We aim to provide a comprehensive explanation of the problem, the steps involved in solving it, and the underlying principles of conditional probability.

Problem Statement

Consider a room containing children from three different grades: first grade, second grade, and third grade. Specifically, there are 3 first-grade children, 5 second-grade children, and 6 third-grade children. If we randomly choose two children from this room, what is the conditional probability of selecting a second-grade child, given that either a first-grade or a third-grade child was selected?

This problem presents a classic conditional probability scenario. We are asked to find the probability of an event (selecting a second-grade child) given that another event (selecting a first-grade or third-grade child) has already occurred. To solve this, we need to carefully consider the total number of ways to select two children, the number of ways to select a second-grade child along with a first-grade or third-grade child, and the number of ways to select a first-grade or third-grade child.

Understanding Conditional Probability

Before diving into the solution, let's clarify the concept of conditional probability. Conditional probability is the probability of an event A occurring given that another event B has already occurred. It is denoted as P(A|B), which reads as "the probability of A given B." Mathematically, conditional probability is defined as:

P(A|B) = P(A ∩ B) / P(B)

where:

  • P(A|B) is the conditional probability of event A given event B.
  • P(A ∩ B) is the probability of both events A and B occurring.
  • P(B) is the probability of event B occurring.

In our problem, event A is selecting a second-grade child, and event B is selecting either a first-grade or a third-grade child. We need to calculate P(A|B), which is the probability of selecting a second-grade child given that a first-grade or third-grade child has already been selected.

Step-by-Step Solution

To solve the problem, we will follow these steps:

  1. Calculate the total number of ways to choose two children.
  2. Calculate the number of ways to choose a first-grade or third-grade child.
  3. Calculate the number of ways to choose a second-grade child and either a first-grade or third-grade child.
  4. Apply the conditional probability formula.

1. Total Number of Ways to Choose Two Children

First, we need to determine the total number of ways to choose two children from the room. There are a total of 3 (first grade) + 5 (second grade) + 6 (third grade) = 14 children. We are choosing 2 children out of 14, which can be calculated using combinations. The number of combinations of choosing k items from a set of n items is given by:

nCk = n! / (k! * (n-k)!)

where:

  • n! (n factorial) is the product of all positive integers up to n.

In our case, n = 14 and k = 2. So, the total number of ways to choose two children is:

14C2 = 14! / (2! * 12!) = (14 * 13) / (2 * 1) = 91

Therefore, there are 91 ways to choose two children from the room.

2. Number of Ways to Choose a First-Grade or Third-Grade Child

Next, we need to find the number of ways to choose two children such that at least one of them is either from the first grade or the third grade. This can be calculated by finding the total number of ways to choose two children and subtracting the number of ways to choose two children from the second grade only.

The number of ways to choose two children from the second grade (5 children) is:

5C2 = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10

So, the number of ways to choose two children such that at least one is from the first or third grade is:

91 (total ways) - 10 (ways to choose two second graders) = 81

Alternatively, we can calculate this directly by considering the following cases:

  • Choosing one first grader and one other child:
    • 3 (first graders) * 11 (non-first graders) = 33 ways
  • Choosing one third grader and one other child:
    • 6 (third graders) * 8 (non-third graders, excluding first graders) = 48 ways
  • Choosing one first grader and one third grader:
    • 3 (first graders) * 6 (third graders) = 18 ways

Adding these up, we get:

33 + 48 - 18 = 63

So, there are 63 ways to select two children where one is from first grade, and 48 ways to select two children where one is from third grade. The overcounting occurs when one child is from first grade and the other is from third grade, which is 3 * 6 = 18 ways. Therefore, using the principle of inclusion-exclusion, the number of ways to select two children such that at least one is from first or third grade is:

(3 * 11) + (6 * 8) + (3 * 6) = 33 + 48 + 18 = 99

This gives the number of ways to select at least one child from either first or third grade. However, we need the number of ways to select exactly one child from either first or third grade. This is where the initial subtraction method provides the correct answer of 91 - 10 = 81.

3. Number of Ways to Choose a Second-Grade Child and Either a First-Grade or Third-Grade Child

Now, we need to find the number of ways to choose two children such that one is from the second grade and the other is from either the first or third grade. We can break this down into two cases:

  • Choosing one second grader and one first grader:
    • 5 (second graders) * 3 (first graders) = 15 ways
  • Choosing one second grader and one third grader:
    • 5 (second graders) * 6 (third graders) = 30 ways

Adding these up, we get:

15 + 30 = 45 ways

So, there are 45 ways to choose a second-grade child and either a first-grade or third-grade child.

4. Apply the Conditional Probability Formula

Finally, we can apply the conditional probability formula:

P(Second Grader | First or Third Grader) = P(Second Grader ∩ (First or Third Grader)) / P(First or Third Grader)

We have already calculated:

  • P(Second Grader ∩ (First or Third Grader)): The number of ways to choose a second-grade child and either a first-grade or third-grade child, divided by the total number of ways to choose two children. This is 45 / 91.
  • P(First or Third Grader): The number of ways to choose two children such that at least one is from the first or third grade, divided by the total number of ways to choose two children. This is 81 / 91.

Therefore, the conditional probability is:

P(Second Grader | First or Third Grader) = (45 / 91) / (81 / 91) = 45 / 81 = 5 / 9

Final Answer

The conditional probability of selecting a second-grade child, given that either a first-grade or third-grade child was selected, is 5/9.

Key Concepts Revisited

This problem effectively illustrates the application of conditional probability in a real-world scenario. By breaking down the problem into smaller steps, we were able to calculate the required probabilities and apply the conditional probability formula. Here are the key concepts we utilized:

  • Combinations: Used to calculate the number of ways to choose items from a set without regard to order.
  • Conditional Probability: The probability of an event occurring given that another event has already occurred.
  • Intersection of Events: The event where both events occur simultaneously.
  • Principle of Inclusion-Exclusion: Used to avoid overcounting when calculating the probability of the union of events.

Practical Implications

Understanding conditional probability has numerous practical applications in various fields, including:

  • Medical Diagnosis: Determining the probability of a disease given certain symptoms.
  • Finance: Assessing the risk of an investment given market conditions.
  • Marketing: Predicting customer behavior based on past actions.
  • Machine Learning: Building predictive models based on conditional probabilities.

By mastering the concept of conditional probability, individuals can make more informed decisions and better understand the relationships between events.

Conclusion

In conclusion, the problem of selecting children from different grades in a room provides a clear and practical example of conditional probability. By following a step-by-step approach and utilizing the appropriate formulas and concepts, we were able to determine the conditional probability of selecting a second-grade child given that a first-grade or third-grade child was selected. This exercise highlights the importance of conditional probability in understanding and predicting real-world events. Whether in mathematics, science, or everyday life, conditional probability serves as a powerful tool for making informed decisions.