Question

\(\mathbf{P . 1 0 8}\) Find \(P(X=2)\) if \(X\) is a binomial random variable with \(n=6\) and \(p=0.3\). if \(X\) is a binomial random variable with \(n=8\) and \(p=0.9\).

Short answer

The probability \(P(X=2)\) for the first condition (\(n=6\) and \(p=0.3\)) will be calculated with the first equation. For the second condition, where \(n=8\) and \(p=0.9\), the probability \(P(X=2)\) will be obtained using the second equation. The results will be the final numeric values obtained from these calculations.

Step-by-step solution

Calculation for first condition

First substitute \(n=6\), \(k=2\), and \(p=0.3\) into the pmf of a binomial distribution: \[P(X=2) = \binom{6}{2}(0.3)^2(1-0.3)^{6-2} = \binom{6}{2}(0.3)^2(0.7)^4\]Calculate this equation to obtain the probability value.

Calculation for second condition

Next, substitute \(n=8\), \(k=2\), and \(p=0.9\) into the pmf of a binomial distribution:\[P(X=2) = \binom{8}{2}(0.9)^2(1-0.9)^{8-2} = \binom{8}{2}(0.9)^2(0.1)^6\]Calculate this equation to obtain the probability value.

Key concepts

Binomial Random Variable

Understanding a binomial random variable is crucial in grasping the foundation of binomial distribution. Simply put, a binomial random variable counts how often a particular event occurs in a fixed number of trials. For instance, if we toss a coin three times, we might count the number of heads that come up. If we define 'getting a head' as our event of interest, and we're looking at three tosses, our binomial random variable could be 0, 1, 2, or 3 - the number of heads that we observe.

In the context of the exercise provided, the number of trials () was set to 6 for the first scenario and 8 for the second. The probability () of success (our event of interest happening, such as getting heads on a coin toss) was 0.3 in the first case and 0.9 in the second. The problem asks you to calculate the probability of getting exactly 2 successes in these scenarios. This is where the binomial distribution comes into play, as it gives us a way to compute these probabilities using a formula specific to binomial random variables.

Probability Mass Function (pmf)

What is pmf?

The probability mass function (pmf) provides the probabilities for discrete random variables, which can take on a finite number of possible outcomes. For binomial random variables, the pmf is an essential tool because it expresses the probability of observing exactly k successes in n trials, given the success probability p for each trial. The pmf of a binomial random variable is given by the formula:
\[P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}\]
where \(P(X=k)\) is the probability of getting k successes, \(n\) represents the number of trials, \(k\) is the number of successes we're interested in, \(p\) is the probability of success on an individual trial, and is the binomial coefficient which calculates the number of ways k successes can occur in n trials.

The step by step solution demonstrates this function in action, using the values from the exercise to calculate the probability of observing exactly 2 successes.

Combinatorics in Probability

Combinatorics, the study of counting, plays a significant role in probability, especially when dealing with binomial distributions. One of its key functions is to determine the number of ways an event can happen. This is fundamentally tied to the concept of the binomial coefficient, often written as \(\binom{n}{k}\), which appears in the pmf formula.

The binomial coefficient \(\binom{n}{k}\) is calculated using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
where \(n!\) (n factorial) is the product of all positive integers up to \(n\), and similarly for \(k!\). This calculation is a combinatorial one, as it counts the number of different ways \(k\) successes can occur during \(n\) independent trials.

In our exercise, the binomial coefficients \(\binom{6}{2}\) and \(\binom{8}{2}\) are used to compute the pmf for the respective scenarios. These coefficients tell us how many ways we can choose 2 successes out of 6 trials and 2 successes out of 8 trials, which are vital figures in finding the final probabilities the problem asks for. Understanding combinatorics is essential to accurately working with binomial distributions and solving problems that involve them.