Problem 34 - Algebra 2

Question

Evaluate \(_{n} C_{x} p^{x} q^{n-x}\) for \(n=7, x=4, p=0.2,\) and \(q=0.8 .\) Round your answer to the nearest thousandth.

Step-by-Step Solution

Answer

The result of evaluating \(_{7} C_{4} * 0.2^{4} * 0.8^{3}\) and rounded to the nearest thousandth is 0.029.

Step 1: Substitution of Variables

Substitute the given values into \(_{n} C_{x} p^{x} q^{n-x}\) to obtain \( _{7} C_{4} * 0.2^{4} * 0.8^{7-4}.\)

Step 2: Simplification

Simplify the formula. Begin by calculating \( _{7} C_{4}\), which stands for the number of combinations of 7 items taken 4 at a time. This can be calculated as \(\frac{7!}{4!(7-4)!}\) which equals 35. Then calculate \(0.2^{4}\) and \(0.8^{3}\) to get 0.0016 and 0.512, respectively. Therefore, the formula becomes \(35 * 0.0016 * 0.512\).

Step 3: Calculation and Rounding

Multiply the numbers together and round to the nearest thousandth. The result is 0.029.

Key Concepts

Combinatorics

Combinatorics is a field of mathematics primarily concerned with counting objects. It helps us determine how many ways we can choose items or arrange them. In our exercise, when faced with \(_{7} C_{4}\), this is a combinatorial expression indicating the number of ways to choose 4 items from 7. The general formula for combinations is given by:

• \(\binom{n}{x} = \frac{n!}{x!(n-x)!}\)

Here, \(n!\) means "n factorial," which is the product of all positive integers up to \(n\). For \(_{7} C_{4}\), we substitute 7 for \(n\) and 4 for \(x\):

• \(\frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)

This tells us there are 35 different ways to choose 4 items from a group of 7. In applications, combinatorics provides a way to calculate probabilities and outcomes in various scenarios.

Probability

Probability is a way of quantifying the likelihood of an event occurring. It ranges from 0 (impossible event) to 1 (certain event). In this problem, the probability concepts are applied to a binomial probability expression. Specifically, we're dealing with two probabilities:

• \(p = 0.2\), the probability of "success" for a single trial.
• \(q = 0.8\), the probability of "failure" (where \(q = 1 - p\)).

In the given formula, each probability is raised to a power which signifies the number of times that outcome occurs across trials.

• \(p^{x} = 0.2^{4}\), meaning the success occurs 4 times.
• \(q^{n-x} = 0.8^{3}\), meaning the failure occurs in the remaining 3 trials (since \(n = 7\) and \(x = 4\)).

These expressions tell us the likelihood of precisely 4 successes in 7 trials, a cornerstone concept of binomial probability.

Binomial Theorem

The binomial theorem expands powers of sums and provides a foundation for the binomial distribution in probability. When you have a series of experiments, like tossing a coin several times, and each experiment leads to one of two outcomes (success or failure), you can often use a binomial distribution to model this.For the binomial probability formula used in our problem, the expression

• \(_{n} C_{x} p^{x} q^{n-x}\)

is crucial. It combines the concepts of combinatorics and probability. Here's a breakdown:

• \(_{n} C_{x}\) calculates the number of ways to achieve a specific number of successes \(x\).
• \(p^{x}\) gives the probability of achieving those \(x\) successes.
• \(q^{n-x}\) provides the probability of the remaining \(n-x\) failures.

Thus, the binomial theorem not only helps with polynomial expansions but also with determining regime details of situations that involve repeated binary trials like our exercise.