Question

Find the chance of throwing at least one ace in a single dice. throw with two \begin{tabular}{rrrrrrr} \multirow{8}{*} {\(\mathrm{S}=\)} & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \end{tabular}

Short answer

The probability of getting at least one ace in a single throw with two dice is \(\frac{11}{36}\).

Step-by-step solution

Identify the sample space

The sample space S consists of all possible outcomes of throwing two dice, which is represented in the given table. Each entry in the table represents the outcome of a throw, with the first number denoting the result of the first dice and the second number the result of the second dice. There are a total of 6x6 = 36 possible outcomes in this sample space.

Identify combinations with at least one ace

To find all combinations with at least one ace, look at the elements in the table where either the first number or the second number is a 1 (i.e., an ace). The highlighted entries are as follows: \begin{tabular}{rrrrrrr} \multirow{8}{*} {\(\mathrm{S}=\)} & \textbf{1,1} & \textbf{1,2} & \textbf{1,3} & \textbf{1,4} & \textbf{1,5} & \textbf{1,6} \\\ & \textbf{2,1} & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\\ & \textbf{3,1} & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\\ & \textbf{4,1} & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\\ & \textbf{5,1} & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\\ & \textbf{6,1} & 6,2 & 6,3 & 6,4 & 6,5 & 6,6 \end{tabular} These choices represent positive outcomes (where there is at least one ace). There are a total of 11 such outcomes.

Calculate probability

To find the probability, divide the number of positive outcomes by the total number of possible outcomes. In this case, there were 11 positive outcomes and 36 total possible outcomes. So, the probability of getting at least one ace in a single throw with two dice is: \(P(\text{at least one ace})= \frac{\text{number of positive outcomes}}{\text{total possible outcomes}} = \frac{11}{36}\)

Key concepts

Sample Space

When we talk about probability, understanding the concept of a sample space is fundamental. It is the set of all possible outcomes that could result from an experiment. For example, if you flip a coin, the sample space is \( S = \{\text{heads}, \text{tails}\} \) because those are the only two outcomes possible.

Now, let's consider the scenario of rolling two six-sided dice. Each die has six faces numbered from 1 to 6, so when throwing two dice, the total number of different outcomes—or the sample space—is the product of the number of outcomes for each die. Mathematically, it's represented as \( 6 \times 6 \) or \( 36 \). This is because each die is independent of the other, allowing for every combination of numbers between the two to occur.

Visualizing this can be done with a grid, as shown in the table, where you see every pair of numbers that can come up (e.g., \( 1,1 \) or \( 6,4 \) and so on). Each cell represents an equally likely outcome, ensuring that the basic premise of sample space—that each outcome is unique and accounted for—is met. This is where our adventure in probability begins: understanding and counting the outcomes in the sample space.

Combinatorics

The next concept we dive into is combinatorics, which is essentially the mathematics of counting and arranging. A critical part of probability is determining how many ways something can happen, and combinatory principles help us calculate this.

In our dice example, we use simple counting methods. We need to find all possible outcomes where at least one die shows an ace (which is the number one). If we roll two dice, there are two ways to get an ace: either the first die shows an ace, the second die shows an ace, or both do.

Possible Outcomes with At Least One Ace

When the first die is an ace, there are six possible outcomes from the second die (1-6). Likewise, when the second die is an ace, the first die also has six possible outcomes. However, we must be careful not to count the outcome where both dice show an ace twice. So we have \( 6 + 6 - 1 = 11 \) unique outcomes where at least one ace appears.

This simple addition and subtraction are basic combinatorial techniques that allow us to avoid overcounting and ensure we include all possible outcomes, laying the groundwork for precise probability calculations.

Probability Calculation

Lastly, we’ll tackle the nitty-gritty of probability calculation. Probability is a measure of how likely an event is to occur out of the number of possible events. In mathematical terms, the probability of an event \( A \) happening is defined as the number of ways \( A \) can occur (which is our favorable outcomes) divided by the total number of possible outcomes (our sample space).

Finding the Probability

To calculate the probability of at least one die showing an ace from a throw of two dice, as established, we have 11 favorable outcomes. Since the number of total possible outcomes in the sample space is 36, we use the formula: \[ P(\text{at least one ace}) = \frac{\text{favorable outcomes}}{\text{total possible outcomes}} = \frac{11}{36} \]

It's important to note that the probability is always a number between 0 and 1, inclusive. A probability of 0 means the event cannot occur, and a probability of 1 means it's certain to occur. Our calculated probability, \( \frac{11}{36} \), tells us that there's a little less than a one-third chance of rolling at least one ace, which means it's possible but not guaranteed. Understanding this process equips students with the ability to tackle a wide array of probability questions moving forward.