The problem describes a scenario where a certain percentage of workers in an industry suffer from a disease. We are asked to find the probability that none of a specific group of 6 workers suffers from this disease.

This situation fits the framework of a binomial distribution because:

• There is a fixed number of trials (n = 6 workers).
• Each trial has only two possible outcomes: a worker suffers from the disease (success) or does not suffer from the disease (failure).
• The probability of success (suffering from the disease) is constant for each worker.
• The trials are independent (whether one worker suffers from the disease does not affect another).

Understanding the Probability of Disease

The incidence of suffering from the disease is given as \(33\frac{1}{3}\%\). Let's convert this percentage into a fraction:

\(33\frac{1}{3}\% = \frac{100}{3}\% = \frac{\frac{100}{3}}{100} = \frac{100}{3 \times 100} = \frac{1}{3}\)

So, the probability that a single worker suffers from the disease is \(p = \frac{1}{3}\).

Calculating the Probability of Not Suffering from Disease

If the probability that a worker suffers from the disease is \(p = \frac{1}{3}\), then the probability that a worker does not suffer from the disease is \(q = 1 - p\).

\(q = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}\)

The probability that a single worker does not suffer from the disease is \(\frac{2}{3}\).

Applying the Binomial Probability Formula

The binomial probability formula for getting exactly \(k\) successes in \(n\) trials is:

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

Where:

• \(n\) is the number of trials (workers), which is 6.
• \(k\) is the number of successes (workers suffering from the disease) we are interested in, which is 0 (no one suffers).
• \(p\) is the probability of success (suffering from the disease) in a single trial, which is \(\frac{1}{3}\).
• \((1-p)\) is the probability of failure (not suffering from the disease) in a single trial, which is \(\frac{2}{3}\).

Calculating the Probability of No Worker Having the Disease

We want to find the probability that exactly 0 out of the 6 workers suffers from the disease. Using the binomial formula with \(n=6\) and \(k=0\):

\(P(X=0) = \binom{6}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^{6-0}\)

Let's calculate each part:

• The binomial coefficient \(\binom{6}{0}\): This represents the number of ways to choose 0 workers out of 6. \(\binom{n}{0} = 1\) for any \(n\). So, \(\binom{6}{0} = 1\).
• The term \(p^k\): This is \(\left(\frac{1}{3}\right)^0\). Any non-zero number raised to the power of 0 is 1. So, \(\left(\frac{1}{3}\right)^0 = 1\).
• The term \((1-p)^{n-k}\): This is \(\left(\frac{2}{3}\right)^{6-0} = \left(\frac{2}{3}\right)^6\).

Now, substitute these values back into the formula:

\(P(X=0) = 1 \times 1 \times \left(\frac{2}{3}\right)^6\)

\(P(X=0) = \left(\frac{2}{3}\right)^6\)

To calculate \(\left(\frac{2}{3}\right)^6\), we raise both the numerator and the denominator to the power of 6:

\(\left(\frac{2}{3}\right)^6 = \frac{2^6}{3^6}\)

Calculate the powers:

• \(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\)
• \(3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729\)

So, the probability is:

\(P(X=0) = \frac{64}{729}\)

This is the probability that none of the 6 workers suffers from the disease.

Summary of Calculation Steps

1. Identify parameters: \(n=6\), disease incidence \(p = 33\frac{1}{3}\% = \frac{1}{3}\).
2. Calculate probability of no disease: \(q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}\).
3. Identify desired outcome: \(k=0\) workers suffer from the disease.
4. Apply binomial probability formula \(P(X=k) = \binom{n}{k} p^k q^{n-k}\).
5. Calculate \(P(X=0) = \binom{6}{0} (\frac{1}{3})^0 (\frac{2}{3})^6\).
6. Simplify: \(1 \times 1 \times (\frac{2}{3})^6 = (\frac{2}{3})^6\).
7. Compute the power: \(\frac{2^6}{3^6} = \frac{64}{729}\).

Revision Table: Key Concepts

Additional Information: Binomial vs. Other Distributions

It's helpful to distinguish the binomial distribution from others when analyzing probability problems involving workers and events like a disease:

• Poisson Distribution: Used for the number of events occurring in a fixed interval of time or space, when events are rare and independent. Not suitable here as we have a fixed number of trials (workers), not an interval.
• Geometric Distribution: Used for the number of trials needed to get the first success. Not suitable here as we are interested in the number of successes within a fixed number of trials, not the trial number of the first success.
• Hypergeometric Distribution: Used when sampling without replacement from a finite population, where the probability of success changes with each trial. Not suitable here as we assume the probability of a worker having the disease is independent and constant across the group of 6.

The binomial distribution is the correct model because we have a fixed sample size (6 workers), each worker either has the disease or not (two outcomes), the probability for each worker is constant (\(\frac{1}{3}\)), and their conditions are independent.

Therefore, the probability that no one out of 6 workers suffers from the disease is \(\frac{64}{729}\).