Understanding the Probability Problem

The problem describes a scenario where we have a fixed number of workers (6) and a known probability that each individual worker suffers from a specific disease (\(33\frac{1}{3}\%\)). We are asked to find the probability that a specific number of these workers (exactly 3) suffer from the disease. This type of problem, where there are a fixed number of independent trials (workers), each with only two possible outcomes (suffers from disease or does not suffer), and a constant probability of success (suffering from disease) for each trial, fits the characteristics of a binomial distribution.

Identifying Binomial Distribution Parameters

In a binomial distribution, we need to identify the following parameters:

• \(n\): The total number of trials. In this case, the total number of workers is 6. So, \(n = 6\).
• \(k\): The number of successful outcomes we are interested in. We want to find the probability that exactly 3 workers suffer from the disease. So, \(k = 3\).
• \(p\): The probability of success on a single trial. The probability that a worker suffers from the disease is given as \(33\frac{1}{3}\%\). Let's convert this to a fraction: \[p = 33\frac{1}{3}\% = \frac{100}{3}\% = \frac{100}{3} \times \frac{1}{100} = \frac{1}{3}\] So, \(p = \frac{1}{3}\).
• \(1-p\): The probability of failure on a single trial (a worker does not suffer from the disease). \[1-p = 1 - \frac{1}{3} = \frac{2}{3}\] So, \(1-p = \frac{2}{3}\).

Applying the Binomial Probability Formula

The probability of getting exactly \(k\) successes in \(n\) trials in a binomial distribution is given by the formula: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\] where \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

Now, we substitute the values we identified: \(n=6\), \(k=3\), \(p=\frac{1}{3}\), and \(1-p=\frac{2}{3}\). \[P(X=3) = \binom{6}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^{6-3}\] \[P(X=3) = \binom{6}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3\]

Calculating the Binomial Coefficient \(\binom{6}{3}\)

First, let's calculate the binomial coefficient \(\binom{6}{3}\): \[\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}\] \[\binom{6}{3} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (3 \times 2 \times 1)}\] \[\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \text{ (Cancelling } 3! \text{ from numerator and denominator)}\] \[\binom{6}{3} = \frac{120}{6}\] \[\binom{6}{3} = 20\]

Calculating the Probabilities \(p^3\) and \((1-p)^3\)

Next, we calculate the terms involving probabilities: \[\left(\frac{1}{3}\right)^3 = \frac{1^3}{3^3} = \frac{1}{27}\] \[\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}\]

Final Probability Calculation for Exactly 3 Workers

Now, we multiply the calculated values together: \[P(X=3) = \binom{6}{3} \times \left(\frac{1}{3}\right)^3 \times \left(\frac{2}{3}\right)^3\] \[P(X=3) = 20 \times \frac{1}{27} \times \frac{8}{27}\] \[P(X=3) = \frac{20 \times 1 \times 8}{27 \times 27}\] \[P(X=3) = \frac{160}{729}\]

Thus, the probability that exactly 3 out of 6 workers suffer from the disease is \(\frac{160}{729}\).

Comparing with Given Options

Let's compare our result with the given options:

Our calculated probability, \(\frac{160}{729}\), matches Option 4.

Revision Table: Key Concepts

Additional Information: When to Use Binomial Distribution

The binomial distribution is applicable when the following conditions are met:

• There is a fixed number of trials (e.g., 6 workers).
• Each trial is independent of the others (the disease status of one worker does not affect another).
• Each trial has only two possible outcomes, typically labeled "success" and "failure" (e.g., suffering from the disease or not suffering).
• The probability of success (\(p\)) is constant for every trial.

In this problem, 'success' is defined as a worker suffering from the disease. The probability of this success is given as \(33\frac{1}{3}\%\) or \(\frac{1}{3}\). We were looking for the probability of exactly 3 successes in 6 trials, which is precisely what the binomial probability formula calculates.