Probability Of Drawing 3 Then 2 Without Replacement

Hey guys! Let's dive into a probability problem that involves drawing cards without replacement. It might sound tricky, but we'll break it down step by step to make it super clear. So, we've got Hiro with his stack of cards, and we need to figure out the chance of him pulling out a 3 first, and then a 2, without putting the cards back in. This is a classic conditional probability question, and we're going to nail it!

Understanding the Problem

The first thing we need to do is really understand what the problem is asking. Hiro has a set of cards, and each card has a number written on it. The numbers are 1, 1, 2, 2, 3, 3, 3, and 4. That means there are eight cards in total. The question we're trying to answer is: what's the probability that Hiro picks a card with a 3 on it first, and then, without putting that card back, picks a card with a 2 on it? This "without replacing" part is super important because it changes the probabilities for the second draw.

To really get this, let's think about why not replacing the card matters. Imagine if Hiro did put the card back after the first draw. The chances of picking a 2 on the second draw would be the same as they were at the beginning. But, because he doesn't replace the card, the total number of cards decreases, and the number of 3s (if he drew a 3) changes. This affects the probability, and that's what makes this problem interesting. We have to consider what happens in the first draw to figure out the chances in the second draw.

Calculating the Probability of Drawing a 3 First

Okay, so let's start with the first part: what's the probability of Hiro drawing a 3 first? This is pretty straightforward. We need to count how many cards have a 3 on them and then divide by the total number of cards. Looking at our set of cards (1, 1, 2, 2, 3, 3, 3, 4), we can see there are three cards with the number 3 on them. And, as we said before, there are eight cards in total.

So, the probability of drawing a 3 first is simply the number of 3s divided by the total number of cards, which is 3/8. This means that if Hiro were to draw a card many, many times, about 3 out of every 8 times, he'd pick a 3. This is our starting point, and now we need to think about what happens after he draws that 3. The next part is where the “without replacement” idea really comes into play, and we have to adjust our thinking to account for the changed situation.

Calculating the Probability of Drawing a 2 Second (Given a 3 Was Drawn First)

Now comes the tricky part, guys! We need to figure out the probability of drawing a 2 after Hiro has already drawn a 3 and not put it back. This is called conditional probability, because the probability of the second event (drawing a 2) depends on what happened in the first event (drawing a 3). So, let's think this through carefully.

After Hiro draws a 3, there are only seven cards left in the stack because he didn't put the 3 back. That's one less card overall. Now, how many cards with a 2 are left? Well, we started with two cards that had a 2 on them (1, 1, 2, 2, 3, 3, 3, 4). Since Hiro only took out a 3, both of those 2s are still in the stack. So, there are still two cards with a 2 on them.

Therefore, the probability of drawing a 2 on the second draw, given that a 3 was drawn first, is the number of 2s left (which is 2) divided by the total number of cards left (which is 7). That gives us a probability of 2/7. Remember, this is different from the probability of drawing a 2 at the beginning because we've changed the situation by taking out a card.

Combining the Probabilities

Alright, we've got the two pieces of the puzzle. We know the probability of drawing a 3 first (3/8), and we know the probability of drawing a 2 second, given that a 3 was drawn first (2/7). Now, how do we combine these to find the overall probability of both events happening? This is where the multiplication rule of probability comes in handy.

The multiplication rule says that if you want to find the probability of two events happening in sequence, you multiply the probability of the first event by the probability of the second event, given that the first event has already occurred. In our case, that means we need to multiply the probability of drawing a 3 first (3/8) by the probability of drawing a 2 second (2/7).

So, the calculation looks like this: (3/8) * (2/7). When we multiply these fractions, we multiply the numerators (the top numbers) and the denominators (the bottom numbers). That gives us (3 * 2) / (8 * 7), which equals 6/56. We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2. That simplifies our answer to 3/28. Therefore, the overall probability of Hiro drawing a 3 first and then a 2 without replacement is 3/28.

Final Answer and Why it Matters

So, after all that calculation, we've arrived at the final answer: the probability of Hiro drawing a 3 first and then a 2 without replacement is 3/28. This might seem like just a number, but it actually tells us something important about the chances of these events happening. If Hiro were to repeat this card-drawing process many times, he would draw a 3 and then a 2 about 3 times out of every 28 attempts. That's a pretty specific outcome, and probability helps us understand how likely it is to occur.

Problems like this are really important in probability because they show us how events can influence each other. The fact that Hiro doesn't replace the card changes the probabilities for the second draw, and we have to take that into account. This concept of conditional probability pops up all over the place in real life, from weather forecasting to medical diagnoses to financial analysis. Understanding how to calculate probabilities in situations where events are dependent on each other is a super valuable skill. Keep practicing these kinds of problems, and you'll become a probability pro in no time!

Let me know if you have any more questions, and happy calculating!