Problem 88 - Precalculus with Limits

Question

In Exercises 85 - 88, consider independent trials of an experiment in which each trial has two possible outcomes: success or failure. The probability of a success on each trial is \( p \), and the probability of a failure is \( q = 1 - p \).In this context, the term \(_nC_kp^kq^{n - k} \) in the expansion of \( \left(p + q\right)^n \) gives the probability of \( k \) successes in the \( n \) trials of the experiment. To find the probability that the sales representative in Exercise 87 makes four sales if the probability of a sale with any one customer is \( \dfrac{1}{2} \), evaluate the term \( _8C_4 \left(\dfrac{1}{2}\right)^4\left(\dfrac{1}{4}\right)^4 \) in the expansion of \( \left(\dfrac{1}{2} + \dfrac{1}{2}\right)^8 \).

Step-by-Step Solution

Answer

The probability that the representative makes exactly four sales is approximately 0.2734.

Step 1: Understand the binomial theorem

The binomial theorem is a way of expanding a binomial expression such as \((x + y)^n\). It states that \((x + y)^n = \sum_{k=0}^{n} {n \choose k} x^{n-k} y^{k}\) where \({n \choose k}\) is read as 'n choose k' and gives the number of ways to choose \(k\) successes from \(n\) trials.

Step 2: Identify the values

The number of trials \(n\) is 8 (given by the sales representative trying with 8 customers), the number of successes \(k\) is 4 (representing making four sales), and the probability of success \(p\) is \(\frac{1}{2}\) (given as the chance of making a sale to a single customer).

Step 3: Substitute in the binomial theorem

We substitute these values in the binomial theorem according to \({n \choose k}p^k(1-p)^{n-k}\). So, \({8 \choose 4}(\frac{1}{2})^{4}(1-\frac{1}{2})^{8 - 4}\) = \(_8C_4\left(\frac{1}{2}\right)^4 \left(\frac{1}{2}\right)^4\).

Step 4: Compute the binomial coefficient

\(_8C_4\), which can also be written as \({8 \choose 4}\), represents the number of ways to choose 4 successes from 8 trials. Calculating this using the combination formula \({n \choose k} = \frac{n!}{k!(n-k)!}\), we get \({8 \choose 4} = \frac{8!}{4!(8-4)!} = 70\).

Step 5: Compute the Probability

Now multiply this with \(\left(\frac{1}{2}\right)^4\) and \(\left(\frac{1}{2}\right)^4\) which are the probabilities of the 4 successes and 4 failures respectively. This gives \(70 \cdot \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^4\) which equals \(0.2734375\).

Key Concepts

Independent Trials

Independent trials are an essential part of probability experiments, especially when dealing with a binomial setting. In this context, each trial has two possible outcomes: success or failure.

Here's what makes trials independent:

• The outcome of one trial does not affect the outcome of another.
• The probability of success remains constant across trials.

In our exercise, the trials are these attempts to make a sale. Each attempt is separate and doesn't influence the others. The probability of success (making a sale) is given as 1/2 for each trial, and this doesn’t change. This setup allows us to apply the binomial formula effectively.

Binomial Coefficient

The binomial coefficient is a key part of evaluating expressions using the binomial theorem. It is often represented as \[ {n \choose k} \] and is read as 'n choose k'. It tells us how many ways we can choose k successes in n trials.

Here's how to calculate it:

• Use the formula \[ {n \choose k} = \frac{n!}{k!(n-k)!} \]
• For n = 8 and k = 4, it becomes \[ \frac{8!}{4! \cdot 4!} \]
• Calculate this to find 70 ways to have exactly 4 successes in 8 trials.

The coefficient is crucial because it quantifies how outcomes can be arranged, an integral part of the binomial probabilities calculation.

Probability Theory

Probability theory underpins the exercise, providing the mathematical framework used in evaluating probabilities of different outcomes.

When considering binomial distributions:

• The probability of success in a single trial is denoted \( p \).
• The probability of failure is \( q = 1 - p \).
• The probability of getting exactly k successes in n independent trials is given by the term \[ {n \choose k}p^kq^{n-k} \].

In the given problem, success (a sale) probability is 1/2, leading to \( p = 1/2 \) and \( q = 1/2 \). For 4 successes in 8 trials, this translates into \[ {8 \choose 4}(\frac{1}{2})^4(\frac{1}{2})^4 \], resulting in a probability of 0.2734375. Understanding these elements allows for quick and effective problem-solving using the binomial theorem.