In this probability experiment, we are dealing with a situation where there are two possible outcomes for each trial: a positive value or a negative value. These outcomes are described as "equally likely," which is a crucial piece of information for setting up our probability model.

Probability Experiment Overview

The problem asks for the probability of obtaining "at most one negative value" in a series of "five trials." This type of scenario, where there's a fixed number of independent trials (five trials), each with two possible outcomes (positive or negative), and the probability of success (getting a negative value) is constant for each trial, perfectly fits the criteria for a Binomial Distribution. In a binomial distribution, we are interested in the number of 'successes' (in this case, negative values) in a given number of trials.

Defining Probability Parameters

Let's define the parameters for our binomial distribution:

• Number of trials (n): We are conducting five trials, so \(n = 5\).
• Probability of success (p): A "success" in this context is obtaining a negative value. Since positive and negative values are equally likely, the probability of obtaining a negative value in any single trial is \(p = \dfrac{1}{2}\).
• Probability of failure (q): A "failure" is obtaining a positive value. Since \(q = 1 - p\), the probability of obtaining a positive value is \(q = 1 - \dfrac{1}{2} = \dfrac{1}{2}\).

The probability mass function (PMF) for a binomial distribution is given by the formula:

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

where:

• \(X\) is the random variable representing the number of successes (negative values).
• \(k\) is the specific number of successes we are interested in.
• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\dfrac{n!}{k!(n-k)!}\), which represents the number of ways to choose \(k\) successes from \(n\) trials.

Calculating Probability of Zero Negative Values

The phrase "at most one negative value" means that the number of negative values obtained can be either zero or one. So, we need to calculate the probability of getting exactly zero negative values, \(P(X=0)\), and the probability of getting exactly one negative value, \(P(X=1)\).

Let's first calculate the probability of obtaining exactly zero negative values (\(k=0\)):

$$P(X=0) = \binom{5}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{5-0}$$

• Calculate the binomial coefficient: \(\binom{5}{0} = \dfrac{5!}{0!(5-0)!} = \dfrac{5!}{0!5!} = 1\).
• Calculate the powers: \(\left(\frac{1}{2}\right)^0 = 1\) and \(\left(\frac{1}{2}\right)^5 = \frac{1^5}{2^5} = \frac{1}{32}\).

Substituting these values back into the formula:

$$P(X=0) = 1 \cdot 1 \cdot \frac{1}{32} = \frac{1}{32}$$

This is the probability of having no negative values among the five trials.

Calculating Probability of One Negative Value

Next, let's calculate the probability of obtaining exactly one negative value (\(k=1\)):

$$P(X=1) = \binom{5}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{5-1}$$

• Calculate the binomial coefficient: \(\binom{5}{1} = \dfrac{5!}{1!(5-1)!} = \dfrac{5!}{1!4!} = \dfrac{5 \times 4 \times 3 \times 2 \times 1}{(1)(4 \times 3 \times 2 \times 1)} = 5\).
• Calculate the powers: \(\left(\frac{1}{2}\right)^1 = \frac{1}{2}\) and \(\left(\frac{1}{2}\right)^4 = \frac{1^4}{2^4} = \frac{1}{16}\).

Substituting these values back into the formula:

$$P(X=1) = 5 \cdot \frac{1}{2} \cdot \frac{1}{16} = \frac{5}{32}$$

This is the probability of having exactly one negative value among the five trials.

Total Probability for At Most One Negative Value

To find the probability of obtaining "at most one negative value," we add the probabilities of getting zero negative values and one negative value:

$$P(X \le 1) = P(X=0) + P(X=1)$$

$$P(X \le 1) = \frac{1}{32} + \frac{5}{32}$$

$$P(X \le 1) = \frac{1+5}{32} = \frac{6}{32}$$

Therefore, the probability of obtaining at most one negative value in five trials is \(\dfrac{6}{32}\).