Solving The Probability Of Drawing Balls And Rolling A Die

Hey guys! Let's dive into a super interesting problem that combines probability, combinatorics, and a little bit of discrete mathematics. It's all about drawing balls from an urn and rolling a die – sounds like fun, right? We're going to break it down step by step so it's crystal clear. So, buckle up and let's get started!

Problem Statement

Okay, so here's the deal. Imagine we have an urn, and inside this urn, there are 5 red balls and 5 blue balls. Now, a boy comes along and he's got a fair die. He rolls this die, and whatever number shows up, that's how many balls he's going to draw from the urn. But here's the catch – he draws them all at once, and he doesn't put any back (that's what we mean by "without replacement").

The big question we need to tackle is to determine the probability of different outcomes. This involves some clever thinking about the possible die rolls, the combinations of balls he could draw, and how to put it all together to calculate those probabilities. So, let's get our thinking caps on and dive in!

Breaking Down the Problem

To really nail this problem, we need to break it down into smaller, more manageable parts. Think of it like building a house – you don't just start throwing bricks together, right? You lay the foundation first. So, let's lay the foundation for this probability problem.

1. Possible Outcomes of the Die Roll

The first thing we need to think about is what can happen when the boy rolls the die. Since it's a fair die, it has six sides, numbered 1 through 6. So, the boy could roll a 1, 2, 3, 4, 5, or 6. Each of these outcomes is equally likely, which is super important for our probability calculations.

2. Number of Balls Drawn

Now, depending on what the die shows, the boy will draw a different number of balls. If he rolls a 1, he draws 1 ball. If he rolls a 2, he draws 2 balls, and so on, all the way up to 6 balls if he rolls a 6. This is a crucial piece of information because the number of balls drawn affects the possible combinations of red and blue balls he could end up with.

3. Combinations of Balls

This is where the combinatorics part comes in. We need to figure out how many different ways the boy can draw a certain number of balls. For example, if he draws 2 balls, he could get 2 red, 2 blue, or 1 red and 1 blue. We need to calculate the number of ways each of these combinations can happen. This is where we'll use combinations (think "n choose k" – you might remember it from math class!).

4. Probability of Each Outcome

Finally, we need to put it all together. We'll calculate the probability of each possible outcome – like drawing 2 red balls, or 3 blue balls, or any other combination. This involves multiplying the probability of rolling a certain number on the die by the probability of drawing a specific combination of balls, given that die roll. It might sound a bit complicated, but we'll break it down step by step.

Combinations: The Heart of the Problem

Let's zoom in on the "combinations of balls" part because it's really the heart of this problem. Understanding combinations is key to figuring out the probabilities. Remember, we're drawing balls without replacement, which means once a ball is drawn, it's not going back into the urn. This affects the probabilities for the subsequent draws.

What are Combinations?

If you've forgotten your combinatorics, don't worry, we'll do a quick refresher. A combination is a way of selecting items from a set where the order doesn't matter. Think of it like picking a team – it doesn't matter if you pick John then Mary or Mary then John; it's the same team.

The formula for combinations is often written as "n choose k" or C(n, k), and it looks like this:

C(n, k) = n! / (k! * (n - k)!)

Where:

• n is the total number of items in the set
• k is the number of items you're choosing
• ! means factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1)

Applying Combinations to Our Problem

In our case, n is the total number of balls in the urn (which is 10 – 5 red and 5 blue), and k is the number of balls the boy draws, which depends on the die roll.

For example, let's say the boy rolls a 3. That means he draws 3 balls. We want to know how many ways he can draw those 3 balls. So, we need to calculate C(10, 3):

C(10, 3) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120

So, there are 120 different ways to draw 3 balls from the urn. But that's not the whole story! We also need to consider the different combinations of red and blue balls.

Red and Blue Combinations

If the boy draws 3 balls, he could get:

• 3 red balls
• 2 red balls and 1 blue ball
• 1 red ball and 2 blue balls
• 3 blue balls

We need to calculate how many ways each of these combinations can happen. This is where it gets a little more detailed, but stick with me!

• 3 red balls: We need to choose 3 red balls out of the 5 red balls in the urn. That's C(5, 3) = 10 ways.
• 2 red balls and 1 blue ball: We need to choose 2 red balls out of 5 (C(5, 2) = 10 ways) and 1 blue ball out of 5 (C(5, 1) = 5 ways). So, the total number of ways is 10 * 5 = 50 ways.
• 1 red ball and 2 blue balls: We need to choose 1 red ball out of 5 (C(5, 1) = 5 ways) and 2 blue balls out of 5 (C(5, 2) = 10 ways). So, the total number of ways is 5 * 10 = 50 ways.
• 3 blue balls: We need to choose 3 blue balls out of the 5 blue balls in the urn. That's C(5, 3) = 10 ways.

See how we're breaking it down? We're using combinations to figure out the different possibilities. Now, let's put this knowledge to work calculating probabilities.

Calculating Probabilities

Okay, we've got the combinations down. Now comes the exciting part – calculating the probabilities! Remember, we need to consider both the probability of rolling a certain number on the die and the probability of drawing a specific combination of balls, given that die roll.

Probability of Rolling the Die

Since it's a fair die, each number (1 through 6) has an equal probability of being rolled. That means the probability of rolling any specific number is 1/6.

Conditional Probability

This is where conditional probability comes in. We're looking for the probability of drawing a certain combination of balls given that the boy rolled a specific number on the die. This is written as P(A | B), which means "the probability of A given B."

In our case, A is the event of drawing a specific combination of balls, and B is the event of rolling a specific number on the die.

To calculate P(A | B), we use the formula:

P(A | B) = P(A and B) / P(B)

But in our case, it's easier to think of it this way: the probability of drawing a certain combination of balls given a die roll is the number of ways that combination can occur divided by the total number of ways to draw that many balls.

Example: Probability of Drawing 2 Red Balls

Let's say we want to find the probability of drawing 2 red balls. This can only happen if the boy rolls a 2 or higher (because he needs to draw at least 2 balls).

Let's break it down for each possible die roll:

• Roll a 2: The boy draws 2 balls. There are C(10, 2) = 45 total ways to draw 2 balls. There are C(5, 2) = 10 ways to draw 2 red balls. So, the probability of drawing 2 red balls given a roll of 2 is 10/45 = 2/9.
• Roll a 3: The boy draws 3 balls. There are C(10, 3) = 120 total ways to draw 3 balls. There are 50 ways to draw 2 red balls and 1 blue ball (as we calculated earlier). So, the probability of drawing 2 red balls and 1 blue ball given a roll of 3 is 50/120 = 5/12.
• Roll a 4: The boy draws 4 balls. We'd need to calculate the number of ways to draw 2 red balls and 2 balls of any color, and so on.
• Roll a 5: The boy draws 5 balls.
• Roll a 6: The boy draws 6 balls.

We'd need to continue these calculations for each possible die roll and each possible combination of balls.

Putting It All Together

To find the overall probability of drawing 2 red balls, we need to consider the probability of each die roll and the conditional probability of drawing 2 red balls given that roll. We use the law of total probability:

P(2 red balls) = P(2 red balls | roll 2) * P(roll 2) + P(2 red balls | roll 3) * P(roll 3) + ...

And so on, for rolls 4, 5, and 6. This gives us a final probability by summing up the probabilities of each possible scenario.

Final Thoughts

This problem is a fantastic example of how probability, combinatorics, and discrete mathematics come together. It might seem a bit daunting at first, but by breaking it down into smaller parts – thinking about the die rolls, the combinations of balls, and the conditional probabilities – we can tackle it step by step.

Remember, the key is to be organized, think clearly, and don't be afraid to get your hands dirty with the calculations. Probability problems can be super rewarding once you crack them! Keep practicing, and you'll become a probability pro in no time. Keep your mind sharp, and thanks for exploring this problem with me!