In An Urn, There Are 10 Spheres (same Size), 3 Of Which Are Green And 7 Are Blue. If 2 Spheres Are Drawn Randomly Without Replacement, What Is The Probability That Both Spheres Selected Are Blue?

Jun 17, 2025 by ADMIN 196 views

Probability of Drawing Two Blue Spheres Without Replacement

In probability theory, understanding the likelihood of events is crucial. This article delves into a classic probability problem: determining the chance of drawing two blue spheres consecutively from an urn containing both blue and green spheres, without replacing the first sphere drawn. This scenario, known as dependent probability, is fundamental in various fields, including statistics, game theory, and risk assessment. The complexity arises from the changing composition of the urn after each draw, influencing the probability of subsequent events. To address this, we'll explore the principles of conditional probability and combinatorial analysis, offering a comprehensive approach to solving this problem. So, let's dive deep into the intricacies of calculating probabilities in scenarios where events are interconnected.

Problem Statement

H2: Understanding the Urn Problem

Consider an urn filled with 10 spheres of identical size. Among these, 3 are green, and 7 are blue. If we randomly select 2 spheres without replacement, what is the probability that both spheres drawn are blue? This problem highlights a situation where the outcome of the first event directly affects the probability of the second event. Initially, the urn contains a known proportion of blue spheres, but this proportion changes after the first sphere is removed. The question requires us to calculate the probability of a compound event – drawing a blue sphere, and then drawing another blue sphere. This is a classic example of a dependent event scenario, where probabilities shift as outcomes occur. Understanding these shifts is key to solving the problem accurately. We will explore the step-by-step approach to this solution, ensuring a clear grasp of how each draw impacts the overall probability.

Solution

H2: Step-by-Step Probability Calculation

To determine the probability of drawing two blue spheres, we break down the problem into sequential events. First, we consider the probability of drawing a blue sphere on the first draw. Since there are 7 blue spheres out of a total of 10, the probability of the first sphere being blue is 7/10. Now, assuming a blue sphere was indeed drawn in the first instance, the situation changes for the second draw. There are now only 6 blue spheres remaining, and the total number of spheres in the urn has decreased to 9. Therefore, the probability of drawing a blue sphere on the second draw, given that a blue sphere was drawn first, is 6/9. To find the overall probability of both events occurring, we multiply the probabilities of each event. This multiplication reflects the combined likelihood of both events happening in sequence. So, the overall probability is (7/10) * (6/9). This calculation embodies the core principle of dealing with probabilities in dependent events: each event's outcome reshapes the probabilities for subsequent events.

H2: Calculating the Combined Probability

Following our step-by-step approach, we've established the probabilities for each draw. The probability of drawing a blue sphere first is 7/10, and the probability of drawing another blue sphere second (given that the first was blue) is 6/9. To find the combined probability of both events happening, we multiply these probabilities together:

(7/10) * (6/9) = 42/90

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 6. This simplification gives us:

42/90 = 7/15

Therefore, the probability of drawing two blue spheres without replacement from the urn is 7/15. This calculation highlights how multiplying probabilities of dependent events gives the likelihood of the entire sequence. The simplified fraction provides a clear and concise representation of the probability, making it easier to interpret and compare.

H2: Alternative Approach: Combinatorial Analysis

Another way to solve this problem is by using combinatorial analysis, which involves counting the number of favorable outcomes and dividing it by the total number of possible outcomes. This method provides a different perspective on the problem, focusing on the counts of different possibilities rather than sequential probabilities. The number of ways to choose 2 blue spheres from 7 is given by the combination formula, often written as "7 choose 2" or C(7, 2). This calculates the number of distinct pairs that can be formed from the 7 blue spheres. Similarly, the total number of ways to choose any 2 spheres from the 10 in the urn is given by C(10, 2). By dividing the number of favorable outcomes (choosing 2 blue spheres) by the total number of outcomes (choosing any 2 spheres), we arrive at the probability. This approach underscores the importance of understanding combinations in probability, particularly in scenarios where the order of selection is not relevant. Let's delve into the calculations involved in this method.

H2: Applying Combinatorial Analysis

In applying combinatorial analysis, we first calculate the number of ways to choose 2 blue spheres out of 7. The combination formula is C(n, k) = n! / (k!(n-k)!), where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. For our case, this is C(7, 2) = 7! / (2!5!) = (7 * 6) / (2 * 1) = 21. This means there are 21 different pairs of blue spheres that could be drawn. Next, we calculate the total number of ways to choose any 2 spheres from the 10 in the urn. This is C(10, 2) = 10! / (2!8!) = (10 * 9) / (2 * 1) = 45. There are 45 possible pairs of spheres that could be drawn. To find the probability, we divide the number of ways to choose 2 blue spheres by the total number of ways to choose 2 spheres: 21/45. This fraction can be simplified by dividing both the numerator and denominator by their greatest common divisor, which is 3. This gives us 7/15, which matches the result obtained using the sequential probability method. This consistency in results reinforces the accuracy of both methods and demonstrates the versatility of probability theory in solving problems.

Conclusion

H2: Summarizing the Probability Solution

In conclusion, the probability of drawing two blue spheres without replacement from an urn containing 3 green and 7 blue spheres is 7/15. We arrived at this answer using two distinct methods: sequential probability calculation and combinatorial analysis. The sequential method involved calculating the probability of drawing a blue sphere on the first draw and then multiplying it by the probability of drawing another blue sphere on the second draw, given that a blue sphere was drawn first. The combinatorial method, on the other hand, focused on counting the number of favorable outcomes (choosing 2 blue spheres) and dividing it by the total number of possible outcomes (choosing any 2 spheres). Both approaches yielded the same result, highlighting the robustness of probability theory in tackling such problems. Understanding these methods is crucial for addressing a wide range of probability-related questions in various real-world scenarios.

This problem exemplifies the fundamental principles of probability, particularly in dependent events. By understanding how to calculate probabilities in situations where outcomes influence subsequent events, we gain valuable insights into decision-making, risk assessment, and statistical analysis. The concepts discussed here form the bedrock of more advanced probability topics, making a solid grasp of these principles essential for anyone venturing further into the field. The ability to solve such problems empowers us to make informed judgments in situations involving uncertainty, a skill that is increasingly valuable in today's data-driven world.