Probability Of Drawing Two Black Balls (Without Replacement)

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Let's dive into a probability problem! Imagine you've got a box filled with colorful balls – specifically, 4 white ones and 6 black ones. Now, you're going to reach in and grab two balls, one after the other, without putting the first one back. The question we're tackling today is: what are the chances that you'll pull out two black balls in a row?

Understanding the Problem

Before we jump into calculations, let's break down what this problem is really asking. We're dealing with conditional probability here, because the outcome of the first draw affects the probabilities in the second draw. Since we're not replacing the first ball, the total number of balls decreases, and the number of black balls also might decrease, depending on what we draw first. That's why understanding conditional probability is important.

Initial Setup

  • Total number of balls: 4 (white) + 6 (black) = 10 balls
  • Number of black balls: 6

Probability of the First Black Ball

The probability of drawing a black ball on your first try is simply the number of black balls divided by the total number of balls. So, we've got:

  • P(First ball is black) = (Number of black balls) / (Total number of balls) = 6 / 10 = 3 / 5 = 0.6 or 60%

So, there's a 60% chance you'll snag a black ball right off the bat. Awesome, right?

Probability of the Second Black Ball (Given the First Was Black)

Okay, here's where it gets a tad trickier. Assuming you did draw a black ball on the first try, things have changed! Now we have:

  • Total number of balls remaining: 9 (since we took one out)
  • Number of black balls remaining: 5 (since we took one black ball out)

So, the probability of drawing a second black ball, given that the first one was black, is:

  • P(Second ball is black | First ball was black) = (Number of remaining black balls) / (Total number of remaining balls) = 5 / 9 ≈ 0.5556 or 55.56%

Calculating the Overall Probability

To find the probability of both events happening (drawing a black ball, then another black ball), we need to multiply the probabilities of each event:

  • P(Two black balls) = P(First ball is black) * P(Second ball is black | First ball was black) = (6 / 10) * (5 / 9) = 30 / 90 = 1 / 3 ≈ 0.3333 or 33.33%

So, the probability of drawing two black balls in a row is approximately 33.33%.

Step-by-Step Breakdown:

To make sure we're all on the same page, let's break down the solution into clear, manageable steps.

  1. Define the events:

    • Event A: Drawing a black ball on the first draw.
    • Event B: Drawing a black ball on the second draw.

  2. Calculate the probability of Event A:

    • P(A) = (Number of black balls) / (Total number of balls) = 6 / 10 = 3 / 5

  3. Calculate the probability of Event B given that Event A has occurred:

    • Since we drew a black ball on the first draw and did not replace it:
      • Number of remaining black balls = 5
      • Total number of remaining balls = 9
    • P(B | A) = (Number of remaining black balls) / (Total number of remaining balls) = 5 / 9

  4. Calculate the probability of both events A and B occurring:

    • P(A and B) = P(A) * P(B | A) = (3 / 5) * (5 / 9) = 15 / 45 = 1 / 3

  5. Convert to percentage (optional):

    • (1 / 3) * 100% ≈ 33.33%

Alternative Approach: Combinations

For those of you who love combinations, here's another way to think about the problem. We can use combinations to directly calculate the number of ways to choose 2 black balls out of the 6 available, and the total number of ways to choose any 2 balls out of the 10.

Combinations Formula

Reminder: The number of combinations of choosing k items from a set of n items is given by:

  • nCk = n! / (k! * (n-k)!) where "!" denotes factorial.

Applying Combinations to Our Problem

  1. Number of ways to choose 2 black balls out of 6:

    • 6C2 = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15

  2. Number of ways to choose any 2 balls out of 10:

    • 10C2 = 10! / (2! * 8!) = (10 * 9) / (2 * 1) = 45

  3. Probability of choosing 2 black balls:

    • P(Two black balls) = (Number of ways to choose 2 black balls) / (Total number of ways to choose 2 balls) = 15 / 45 = 1 / 3 ≈ 0.3333 or 33.33%

As you can see, both methods lead to the same answer. Choosing the right approach sometimes depend on the context.

Key Concepts Revisited

Let's make sure we've nailed down the core concepts used in solving this probability puzzle.

Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has already occurred. It's written as P(A | B), which reads as "the probability of A given B." In our problem, we needed to calculate the probability of drawing a second black ball, given that the first ball drawn was black. This is a classic example of conditional probability because the first event changes the conditions for the second event.

Independent vs. Dependent Events

It's also crucial to understand the difference between independent and dependent events.

  • Independent Events: The outcome of one event does not affect the outcome of the other event. For example, flipping a coin twice – the result of the first flip doesn't change the probabilities of the second flip.
  • Dependent Events: The outcome of one event does affect the outcome of the other event. Our ball-drawing problem is a perfect example of dependent events because removing the first ball changes the composition of the remaining balls.

If we had replaced the first ball, the events would have been independent, and the probability of drawing a black ball on the second try would have remained the same as the first try (6/10). But since we didn't replace the ball, the probabilities shifted, making it a dependent event scenario.

Combinations

Combinations are a way to count the number of possible selections of items from a set, where the order of selection doesn't matter. The formula for combinations helps us determine how many different ways we can choose k items from a set of n items without regard to the order.

Real-World Applications

So, why bother learning about drawing balls from a box? Well, these probability concepts show up in all sorts of real-world situations!

  • Quality Control: Imagine a factory producing light bulbs. Quality control engineers might randomly select a few bulbs to test for defects. They use probability to estimate the likelihood of finding defective bulbs in a larger batch.
  • Medical Testing: When doctors order medical tests, they use probability to interpret the results. For example, the probability of a positive test result accurately indicating a disease, taking into account the test's sensitivity and specificity, and the prevalence of the disease in the population.
  • Finance and Investing: Investors use probability to assess the risk and potential returns of different investments. They might analyze historical data to estimate the probability of a stock price increasing or decreasing.
  • Insurance: Insurance companies rely heavily on probability to calculate premiums. They assess the likelihood of various events occurring (like accidents, illnesses, or natural disasters) to determine how much to charge for insurance coverage.
  • Games of Chance: Of course, probability is at the heart of games like poker, blackjack, and lotteries. Understanding probability can help you make more informed decisions (though it doesn't guarantee you'll win!).

Conclusion

So, there you have it! The probability of drawing two black balls in a row from a box containing 4 white balls and 6 black balls, without replacement, is 1/3 or approximately 33.33%. We arrived at this answer using both conditional probability and combinations, illustrating how different approaches can be used to tackle the same problem. Understanding these concepts isn't just about solving puzzles; it's about building a foundation for making informed decisions in a world filled with uncertainty. Keep practicing, and you'll be a probability pro in no time!