Probability Of Drawing A White Ball: A Step-by-Step Guide
Hey everyone! Today, we're diving into a classic probability problem. Imagine this: You've got an urn filled with four white balls and six black balls. Now, you reach in and pull out two balls without putting the first one back in. The big question is: What are the chances that at least one of those balls you grabbed is white? Let's break this down, step by step, so it's super clear.
Understanding the Basics: Probability and Combinations
Before we jump into the problem, let's get our heads around a couple of key concepts. First up, probability. Probability is all about figuring out the chances of something happening. We express it as a number between 0 and 1 (or as a percentage between 0% and 100%). A probability of 0 means it's impossible, and a probability of 1 means it's a sure thing. In our case, we're trying to figure out the probability of getting at least one white ball. This means we are finding the likelihood of getting one white ball or two white balls. To calculate this probability, we'll need to use some basic principles of probability.
Now, let’s talk about combinations. A combination is a way of selecting items from a larger group where the order doesn't matter. Think of it like picking a team of people. It doesn't matter if you pick John then Sarah or Sarah then John – the team is the same. The formula for combinations is:
C(n, k) = n! / (k! * (n-k)!)
Where:
n = the total number of items k = the number of items you're choosing ! means factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1)
We'll use this formula to figure out how many different ways we can draw balls from the urn.
Now to calculate the probability, you need to consider the events that satisfy the condition of “at least one white ball”. This means we will analyze three possible scenarios: a white ball is drawn in the first extraction and a white ball is drawn in the second extraction; a white ball is drawn in the first extraction and a black ball in the second extraction; and a black ball is drawn in the first extraction and a white ball in the second extraction. The simplest way to calculate the probability of the original statement is to calculate the probability of the complementary event (in which the event does not happen), and subtract it from 1. The complementary event in this case is to draw two black balls, in other words, to not draw a white ball. Let's get started.
Why Combinations are Important Here
We're using combinations because we don't care about the order in which we draw the balls. Getting a white ball then a black ball is the same for our purposes as getting a black ball then a white ball – as long as we have at least one white ball. This simplifies our calculations significantly, making it easier to see all the possible outcomes.
Calculating the Probability: The Step-by-Step Approach
Alright, let's get down to the nitty-gritty and calculate the probability of getting at least one white ball. We can solve this problem by considering the complement of the event. The complement of the event of getting at least one white ball is getting no white balls at all, i.e., getting two black balls. So, let’s calculate the probability of the complement and subtract it from 1.
Step 1: Calculate the total number of ways to draw two balls.
We have a total of 10 balls (4 white + 6 black). We want to choose 2. Using the combination formula:
C(10, 2) = 10! / (2! * 8!) = (10 * 9) / (2 * 1) = 45. There are 45 different ways to draw two balls from the urn. Remember, the total number of ways to pick any two balls from the urn, regardless of color is 45. That is, 45 possible scenarios.
Step 2: Calculate the number of ways to draw two black balls.
We have 6 black balls, and we want to choose 2. Using the combination formula:
C(6, 2) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15. There are 15 ways to draw two black balls.
Step 3: Calculate the probability of drawing two black balls.
Probability (two black balls) = (Number of ways to draw two black balls) / (Total number of ways to draw two balls) = 15 / 45 = 1/3.
Step 4: Calculate the probability of drawing at least one white ball.
Since we calculated the probability of the event “not getting at least one white ball”, which is to draw two black balls, the probability of at least one white ball is the complementary of that probability.
Probability (at least one white ball) = 1 - Probability (two black balls) = 1 - (1/3) = 2/3. So, the probability of drawing at least one white ball is 2/3, or approximately 66.67%.
Alternative Approach: Direct Calculation
We can also directly calculate the probability. Let's find the probability of getting exactly one white ball and the probability of getting two white balls, and then add those probabilities together.
One White Ball and One Black Ball
- Ways to get one white ball: 4 (ways to choose 1 white ball) * 6 (ways to choose 1 black ball) = 24.
- Probability: 24/45
Two White Balls
- Ways to get two white balls: C(4, 2) = (43)/(21) = 6
- Probability: 6/45
Total Probability (Direct Method)
Probability (at least one white ball) = P(one white, one black) + P(two white) = 24/45 + 6/45 = 30/45 = 2/3.
See? We get the same answer! Both methods confirm that the probability of drawing at least one white ball is 2/3.
Breaking Down the Calculation: Key Considerations
Let's unpack a few things to make sure we're all on the same page. The key to this problem is understanding that we're dealing with events that happen without replacement. That's super important.
Without Replacement
When we say