Question

One of the numbers \(1,2, \ldots, 6\) is to be chosen by casting an unbiased die. Let this random experiment be repeated five independent times. Let the random variable \(X_{1}\) be the number of terminations in the set \(\\{x: x=1,2,3\\}\) and let the random variable \(X_{2}\) be the number of terminations in the set \(\\{x: x=4,5\\}\). Compute \(P\left(X_{1}=2, X_{2}=1\right)\)

Short answer

The calculated probability when the random variable \(X_{1}\) is equals to 2 and \(X_{2}\) equals to 1 in a random experiment involving five independent throws of an unbiased die is 0.1389

Step-by-step solution

Determine the Total Number of Possible Outcomes

When the experiment is performed five times, we are basically doing 5 independent throws with an unbiased die. Since there are 6 possible outcomes on each throw, there are \(6^5\) or 7776 possible outcomes.

Define the Rules of the Experiment

We are looking for the instances where 2 numbers fall between 1 and 3 and 1 number falls between 4 and 5. This means that in 5 throws, 2 are in the set {1, 2, 3}, 1 in the set {4, 5} and the remaining 2 can be any number between 1 and 6.

Calculate the Number of Desired Outcomes

The number of ways we can pick 2 out of 5 falls (for {1,2,3}) can be calculated using binomial coefficient as _5C_2 = 10. The number of ways to pick 1 out of the remaining 3 falls (for {4, 5}) is _3C_1 = 3. And, remaining 2 falls can be any number between [1,6] which is \(6^2=36\). Hence, total desired outcomes is product of all three which is 10*3*36 = 1080.

Calculate the Probability

With this information, we can calculate the probability as \(P(X_{1}=2, X_{2}=1)=\ \frac {number\ of\ desired\ outcomes}{total\ number\ of\ outcomes} = \frac {1080}{7776} = 0.1389\). We found this probability by counting the number of desired outcomes and then divided by the total number of outcomes.

Key concepts

Random Variables

In the world of probability and statistics, a random variable is a numerical description of the outcome of a statistical experiment. A random variable thus assigns a number to each potential outcome of an experiment. It's important to note that these variables can be discrete or continuous. In the case of rolling a die, as mentioned in our example exercise, the random variables are discrete.

In this particular problem, we are considering two random variables, denoted as \(X_{1}\) and \(X_{2}\). The first, \(X_{1}\), corresponds to the number of times a certain outcome (a roll of 1, 2, or 3) occurs, while the second, \(X_{2}\), relates to a different outcome (a roll of 4 or 5). To compute probabilities involving these random variables, we combine our understanding of the die's behavior with combinatorial calculations. We are effectively asking: 'What is the chance that in five throws, we'll see our specific combination of outcomes related to \(X_{1}\) and \(X_{2}\)?' This process is a foundation of probability theory and is pivotal for analyzing a variety of real-world scenarios.

Binomial Coefficient

The binomial coefficient, often pronounced as 'n choose k', is a fundamental concept in combinatorics. It calculates the number of ways we can choose \(k\) objects from a set of \(n\) distinct objects without considering the order of selection. Mathematically, it's represented by the notation \(_nC_k\) or \(\binom{n}{k}\).

Understanding the binomial coefficient is essential for solving problems involving probability where there are a certain number of successes in a series of independent trials, just like in our die-rolling experiment. For example, choosing 2 out of 5 rolls to land a number between 1 and 3 is calculated using the binomial coefficient formula: \(\binom{5}{2} = 10\). This tells us there are 10 different ways that two rolls could result in 1, 2, or 3 while disregarding the other three rolls. Similarly, choosing one outcome out of three rolls to get a number between 4 and 5 gives us \(_3C_1 = 3\), indicating three possible ways.

Probability Calculation

To calculate the probability of a specific outcome in a random experiment, we need to know two things: the number of desired outcomes and the total number of possible outcomes. The probability is then found by dividing the number of desired outcomes by the total number of possible outcomes.

Let's apply these concepts to our dice exercise. The total possible outcomes are the number of different results we could get from rolling a die five times independently, which is \(6^5\) (since a die has 6 faces and each roll is independent). We've previously determined the number of desired outcomes, by considering our restrictions on \(X_{1}\) and \(X_{2}\), to be 1080. Therefore, to find the probability of our event, we divide the number of desired outcomes by the total number of possible outcomes, yielding a probability of \(\frac{1080}{7776}\), which simplifies to 0.1389 or 13.89%.

Remember, this process is not just about calculation—it's about conceptualizing the experiment, understanding the possible outcomes, counting them correctly, and applying combinatorial skills to find the probability of events.