Rolling A Six-Sided Die Probability Of 5 Then 3

Hey guys! Let's dive into a probability problem that involves rolling a six-sided die, also known as a number cube, not just once, but twice! Our mission is to figure out the probability of rolling a 5 on the first roll and then rolling a 3 on the second roll. To solve this, we will explore the fundamental principles of probability and how they apply to independent events.

Understanding Probability Basics

First, probability in its simplest form is a way to measure how likely something is to happen. We often express probability as a fraction, where the numerator (the top number) represents the number of favorable outcomes, and the denominator (the bottom number) represents the total number of possible outcomes. So, if we're talking about rolling a standard six-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Each of these outcomes is equally likely, assuming we have a fair die. Therefore, the probability of rolling any specific number on a single roll is 1 (the favorable outcome) divided by 6 (the total possible outcomes), which equals $\frac{1}{6}$. This is a core concept to grasp as we move forward, guys, because it sets the stage for calculating probabilities in more complex scenarios, such as rolling the die multiple times.

When we consider rolling the die twice, we're dealing with what we call independent events. Independent events are events where the outcome of one doesn't affect the outcome of the other. Think about it: the result of your first roll has absolutely no impact on what you'll get on your second roll. Each roll is a fresh start, a new chance. This independence is crucial because it allows us to calculate the combined probability of both events occurring by simply multiplying their individual probabilities. This rule of multiplication for independent events is a cornerstone of probability theory, and it's what helps us tackle problems like the one we're faced with here. We're not just looking at one roll in isolation; we're considering the sequence of two rolls and the likelihood of a specific sequence (rolling a 5, then a 3) occurring. To get a handle on this, we need to break down each roll separately and then bring them together using this multiplication rule. Keep this in mind, because we're about to put this concept to work!

Calculating the Probability of Rolling a 5

The first part of our mission is to determine the probability of rolling a 5 on the first roll of the six-sided die. Remember, we've already laid the groundwork by understanding that probability is about favorable outcomes versus total possible outcomes. In this scenario, we're looking for one specific outcome: rolling a 5. Since our die has six sides, numbered 1 through 6, there is only one side that shows a 5. This means we have one favorable outcome.

Now, let's think about the total possible outcomes. As we've established, there are six sides to the die, each representing a different number. So, when you roll the die, there are six equally likely possibilities: you could roll a 1, 2, 3, 4, 5, or 6. These are your total possible outcomes. This is a straightforward situation where every outcome is equally probable, making our calculation quite direct. We don't have to worry about any side being more likely to land face up than any other, which simplifies our task considerably. With the favorable outcome and total possible outcomes clearly defined, we're ready to express this as a probability.

To express the probability, we create a fraction. The number of favorable outcomes (rolling a 5) goes on top, as the numerator, and the total number of possible outcomes (the six sides of the die) goes on the bottom, as the denominator. So, the probability of rolling a 5 on the first roll is $\frac{1}{6}$. This fraction tells us that, on any given roll, there is a one in six chance of landing on a 5. It's a small chance, but it's a precise way to quantify the likelihood of this specific event happening. We've now successfully pinned down the probability of the first part of our sequence, and we're ready to move on to the second roll, where we're aiming for a 3. Keep this fraction $\frac{1}{6}$ in mind, guys, because it's going to be a crucial piece of our final calculation.

Determining the Probability of Rolling a 3

Next up, we need to figure out the probability of rolling a 3 on the second roll. Just like with the 5, we're focusing on a specific outcome, but this time it's the number 3. The principles here are exactly the same as before, because each roll of the die is an independent event. This means that whatever happened on the first roll doesn't change the odds for the second roll. It's a fresh start, and the die doesn't have any memory of previous rolls. This independence is what allows us to treat each roll as its own separate probability calculation.

When we think about favorable outcomes for rolling a 3, we realize there's only one side of the die that has the number 3 on it. So, just as with the 5, we have one favorable outcome. This might seem simple, but it's important to be precise about identifying what we're looking for. In this case, we're not interested in any other number; we're specifically targeting the 3. Recognizing that we have one single favorable outcome is a key step in setting up our probability calculation correctly.

As for the total possible outcomes, these remain the same as they were for the first roll. Our six-sided die still has six sides, each with a different number from 1 to 6. So, there are still six equally likely possibilities for the second roll. This consistency is what makes working with standard dice so predictable and allows us to apply basic probability rules with confidence. Now that we've identified both the favorable outcome and the total possible outcomes for rolling a 3, we're ready to express this as a probability, just as we did for rolling a 5. This will give us the second piece of the puzzle we need to solve the overall probability of rolling a 5 and then a 3.

To express the probability of rolling a 3, we again form a fraction. The favorable outcome (rolling a 3) goes on top as the numerator, and the total possible outcomes (the six sides of the die) go on the bottom as the denominator. This gives us a probability of $\frac{1}{6}$ for rolling a 3 on the second roll. Just like the probability of rolling a 5, this means there's a one in six chance of rolling a 3 on any given roll of the die. Now we have the individual probabilities for each roll, and we're ready to combine them to find the probability of the entire sequence. Remember this $\frac{1}{6}$, guys, because it's the other critical piece we need to put it all together.

Combining Probabilities for Independent Events

Now, the crucial step is to combine the individual probabilities to find the overall probability of rolling a 5 and then a 3. Remember, we're dealing with independent events here, meaning the outcome of the first roll doesn't impact the outcome of the second roll. This is key because it allows us to use a simple rule: to find the probability of two independent events both happening, you multiply their individual probabilities together. This rule is a cornerstone of probability theory, and it makes calculating the likelihood of sequences of events manageable.

We've already determined that the probability of rolling a 5 on the first roll is $\frac{1}{6}$, and the probability of rolling a 3 on the second roll is also $\frac{1}{6}$. So, to find the probability of rolling a 5 and then a 3, we simply multiply these two fractions together. This is where our earlier work pays off, guys, because we've broken the problem down into its component parts and calculated each probability separately. Now, we just need to apply this multiplication rule to get our final answer.

When multiplying fractions, you multiply the numerators (the top numbers) together to get the new numerator, and you multiply the denominators (the bottom numbers) together to get the new denominator. So, in this case, we're multiplying $\frac{1}{6}$ by $\frac{1}{6}$. This means we multiply 1 by 1 for the numerator, which gives us 1, and we multiply 6 by 6 for the denominator, which gives us 36. This calculation is straightforward, but it's important to get the mechanics right to ensure we arrive at the correct final probability. The result of this multiplication is the probability of the entire sequence of events occurring, and it's the answer we've been working towards.

Calculating the Final Probability

Let's perform the multiplication: $\frac{1}{6} imes \frac{1}{6} = \frac{1 imes 1}{6 imes 6} = \frac{1}{36}$. So, the final probability of rolling a 5 on the first roll and a 3 on the second roll is $\frac{1}{36}$. This fraction represents the likelihood of this specific sequence of events occurring when you roll a six-sided die twice. It tells us that, out of all the possible outcomes you could get when rolling the die twice, only one of those outcomes is the sequence we're interested in (rolling a 5, then a 3).

This result highlights that while each individual roll has a probability of $\frac{1}{6}$ for a specific number, the probability of getting a specific sequence of numbers is lower. This is because we're considering the intersection of two independent events, and the more events you combine, the lower the probability of the entire sequence typically becomes. The fraction $\frac{1}{36}$ is a small probability, indicating that this particular sequence is not very likely to occur by chance. This is a key insight in probability: combining events often reduces the likelihood of a specific outcome.

Now, let's connect this back to the original question. We were asked to identify the expression that can be used to find $P(5, \text{then } 3)$, which is the probability of rolling a 5, then a 3. We've walked through the entire process, breaking down the problem into individual probabilities and then combining them. The expression we used was $\frac{1}{6} imes \frac{1}{6}$, which accurately represents the multiplication of the individual probabilities. This expression is the correct way to represent the calculation we performed, and it directly leads to the final probability of $\frac{1}{36}$. So, we've not only solved the problem but also identified the correct expression to use in this type of scenario. Great job, guys!

Identifying the Correct Expression

Therefore, the correct expression to find $P(5, \text{then } 3)$ is $\frac{1}{6} imes \frac{1}{6}$. This expression accurately represents the probability of rolling a 5 on the first roll ($\frac{1}{6}$) and then rolling a 3 on the second roll ($\frac{1}{6}$), with the multiplication symbolizing the combination of these two independent events. We've walked through the entire problem, from understanding basic probability principles to calculating individual probabilities and then combining them to find the overall probability. This process not only gives us the answer but also a solid understanding of why it's the answer. You guys nailed it!

In summary, we tackled a probability problem involving rolling a six-sided die twice. We learned that the probability of rolling a 5 and then a 3 is calculated by multiplying the individual probabilities of each event. Since each roll is an independent event with a probability of $\frac{1}{6}$ for any specific number, the combined probability is $\frac{1}{6} imes \frac{1}{6} = \frac{1}{36}$. The correct expression to represent this calculation is indeed $\frac{1}{6} imes \frac{1}{6}$. I hope this explanation helps you grasp the concept of calculating probabilities for independent events. Keep practicing, and you'll become probability pros in no time, guys!