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 joint probability \(P\left(X_{1}=2, X_{2}=1\right)\) is obtained by multiplying the binomial probabilities calculated for \(X_{1}=2\), and \(X_{2}=1\) respectively.

Step-by-step solution

Computing probability associated with random variable \(X_{1}\)

Here, \(x_{1}\) can take the values 1, 2 or 3. So, \(p_{1} = P(x_{1} = 1, 2, 3) = 3/6 = 0.5\). Now, the number of times this occurs in 5 trials (let's call it \(k_{1}\)) is given as 2. Thus, the probability associated with \(X_{1}\) in 5 trials can be calculated using the binomial distribution formula: \[P(X_{1} = 2) = \tbinom{5}{2} \times (0.5)^2 \times (1-0.5)^{5-2}\]

Computing probability associated with random variable \(X_{2}\)

Here \(x_{2}\) can take the values 4 or 5. So, \(p_2 = P(x_2 = 4, 5) = 2/6 = 1/3\). Now, the number of times this occurs in 5 trials (let us call it \(k_{2}\)) is given as 1. Thus, the probability associated with \(X_{2}\) in 5 trials can be calculated using the binomial distribution formula: \[P(X_{2} = 1) = \tbinom{5}{1} \times (1/3)^1 \times (1 - 1/3)^{5-1}\]

Calculating the required joint probability

From the rule of multiplication for independent events, the required joint probability is \(P\left(X_{1}=2, X_{2}=1\right) = P(X_{1}=2) \times P(X_{2}=1)\). Substitute the calculated probabilities from Step 1 and Step 2 to get the output.

Key concepts

Binomial Distribution

Binomial distribution is a cornerstone concept in mathematical statistics fundamental to understanding how to model scenarios involving a fixed number of independent trials with only two possible outcomes: success or failure. Take, for example, a coin toss: if you flip a fair coin multiple times, the number of heads (success) you get follows a binomial distribution.

Returning to the exercise, when a die is cast, landing on 1, 2, or 3 is considered 'success' for the random variable \(X_{1}\), and landing on 4 or 5 is a 'success' for \(X_{2}\). The trials are independent, which means the outcome of one throw doesn’t affect the others. Here, binomial distribution is used to calculate the probability of seeing a certain number of 'successes' across multiple trials.

For a given number of trials (n) and the probability of success in a single trial (p), the formula for binomial distribution is \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(X\) is the number of successes, and \(k\) is the specific number of successes out of n trials that we’re calculating the probability for.

Joint Probability

Joint probability is used to measure the likelihood of two events occurring at the same time. In the context of statistical problems that involve random variables, like our die-casting exercise, it accounts for the probability of intersection between the occurrences of separate events.

In the given problem, we are looking for the joint probability of \(X_{1}=2\) and \(X_{2}=1\). These are independent events because the result of one set of die rolls doesn't influence the others. Since the events are mutually exclusive, meaning \(X_{1}\) and \(X_{2}\) do not overlap in outcomes, we can multiply their individual probabilities to find the joint probability: \[P(X_{1}=2, X_{2}=1) = P(X_{1}=2) \times P(X_{2}=1)\]
Calculating joint probability is essential in statistics as it helps in understanding the relation and interaction between different events or random variables.

Random Variables

In statistics, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. There are two types of random variables: discrete, which take on a countable number of distinct values (like the outcome of a die roll), and continuous, which take on values within a range or interval.

The variables \(X_{1}\) and \(X_{2}\) in our exercise are discrete random variables. \(X_{1}\) represents the number of die rolls resulting in 1, 2, or 3, while \(X_{2}\) counts how many times 4 or 5 appears. The work we do to calculate probabilities for these variables involves understanding not just their individual distributions but also how they might interact, which helps in solving more complex problems involving multiple random variables.

Understanding random variables is crucial, as they are the foundation for probability distributions—a key concept in statistics, which helps in predicting outcomes and making sense of random events.