Question

Consider a binomial random variable with \(n=9\) and \(p=.3 .\) Let \(x\) be the number of successes in the sample. Evaluate the probabilities in Exercises \(7-10 .\) The probability that \(x\) is exactly 2 .

Short answer

Answer: The probability of exactly 2 successes in this binomial experiment is approximately 0.2668.

Step-by-step solution

Recall the binomial probability formula

The binomial probability formula for calculating the probability of exactly \(x\) successes in \(n\) trials is given by: \(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\), where \(0 \leq k \leq n\) Here, \(P(X=k)\) is the probability of \(k\) successes, \(\binom{n}{k}\) is the binomial coefficient or the number of ways to choose \(k\) successes from \(n\) trials, \(p\) is the probability of success, and \(1-p\) is the probability of failure.

Plug in the values

We are given \(n=9\), \(p=0.3\), and \(x=2\). Plugging these values into the binomial probability formula, we get: \(P(X=2)=\binom{9}{2}(0.3)^2(1-0.3)^{9-2}\)

Calculate the binomial coefficient

First, we need to calculate the binomial coefficient \(\binom{9}{2}\). The binomial coefficient is given by: \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\), where \(n!\) is the factorial of \(n\). So, \(\binom{9}{2}=\frac{9!}{2!(9-2)!}=\frac{9!}{2!7!}=\frac{9\times8}{2\times1}=36\). Now, plugging the value of the binomial coefficient back into the formula, we get: \(P(X=2)=36(0.3)^2(1-0.3)^{9-2}\)

Evaluate the probability

Now, we can compute the probability as follows: \(P(X=2)=36(0.3)^2(0.7)^7\approx 0.2668\) Therefore, the probability that \(x\) is exactly 2 is approximately 0.2668.

Key concepts

Binomial Distribution

The binomial distribution is a probability distribution that summarizes the likelihood that a variable will take one of two independent states across several trials or instances. For example, it can be used to determine the probability of getting a certain number of heads in a series of coin tosses.

In the provided exercise, we're examining a scenario where the binomial distribution helps us find out the probability of a certain number of successes occurring in a set number of trials. Each 'trial' is an independent event with two possible outcomes - success or failure - and the probability of success remains consistent across each trial.

Because the binomial distribution can predict the probabilities of various possible numbers of successes, it has a wide range of applications, from quality control in manufacturing to predicting the outcomes of sports games.

Binomial Random Variable

A binomial random variable is the number of successes in a series of independent 'yes or no' experiments, or Bernoulli trials, each with the same probability of success. For instance, if we consider tossing a fair coin 9 times, each toss is a Bernoulli trial with two possible outcomes: heads or tails. If we define success as getting heads, the number of heads we get after those 9 tosses is the binomial random variable.

In our exercise, the binomial random variable represents the number of successes in the 9 trials, which is denoted by the letter 'x'. The trials are considered to be independent, meaning the outcome of one trial does not influence the outcomes of other trials.

Probability of Success

The probability of success, often represented by 'p', is the chance of a single trial resulting in a success. This probability must remain constant across all trials for the distribution to be binomial. In our example, the probability of success is set at 0.3, or 30%, which means each of the 9 trials has a 30% chance of being successful.

It's crucial to differentiate between the probability of a single event and the overall probability of a series of events. Even though each event has the same probability of success, the probability of achieving a certain number of successes across multiple trials is a separate calculation, which leads us to our final concept - probability calculation.

Probability Calculation

Calculating the probability of a certain number of successes over multiple trials in a binomial distribution involves several steps. The formula used for calculating the probability of exactly 'k' successes in 'n' trials is:

\(P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\)

Here's a quick rundown of the calculations we did in the solution step-by-step:

• We determined the number of successes we were looking for ('k', which in the exercise was 2).
• We used the binomial probability formula by plugging in the values for 'n', 'k', and 'p'.
• We calculated the binomial coefficient, which represents the number of ways 'k' successes can occur in 'n' trials.
• We then multiplied the binomial coefficient by the probability raised to the power of the number of successes and the probability of failure raised to the number of non-successes.
• Through these calculations, we found that the probability of getting exactly 2 successes in 9 trials with a success probability of 30% is approximately 0.2668.

This probability calculation is crucial for understanding patterns of success in not only academic exercises but also real-world scenarios.