Understanding the Probability Problem

This question asks us to find the probability that a certain number of workers (4 or more) out of a group of 6 will suffer from a specific disease. We are given the probability that any single worker might suffer from this disease.

Key information provided:

• The probability of a single worker suffering from the disease, denoted as \( p \), is 20% or 0.20.
• The total number of workers chosen at random, denoted as \( n \), is 6.
• We need to find the probability that 4 or more workers suffer from the disease. This means we are interested in the cases where exactly 4 workers, exactly 5 workers, or exactly 6 workers suffer from the disease.

Applying Binomial Distribution

This scenario fits the Binomial distribution criteria because:

• There are 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 (failure).
• The probability of success ( \( p=0.20 \) ) is the same for each worker.
• The trials are independent; one worker suffering does not affect another's chance.

The probability of failure (a worker not suffering) is \( q = 1 - p = 1 - 0.20 = 0.80 \).

The Binomial probability formula is used to calculate the probability of getting exactly \( k \) successes in \( n \) trials:

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

Where:

• \( \binom{n}{k} \) is the number of combinations of choosing \( k \) items from a set of \( n \), calculated as \( \frac{n!}{k!(n-k)!} \).
• \( p^k \) is the probability of \( k \) successes.
• \( q^{n-k} \) is the probability of \( n-k \) failures.

Calculating the Probability

We need to calculate the probability for 4, 5, and 6 workers suffering from the disease and sum them up. That is, we need to find \( P(X \ge 4) = P(X=4) + P(X=5) + P(X=6) \).

Probability for Exactly 4 Workers (P(X=4))

Here, \( n=6 \), \( k=4 \), \( p = \frac{1}{5} \), \( q = \frac{4}{5} \).

• Calculate the combination: \( \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5}{2 \times 1} = 15 \).
• Calculate the probability: \( P(X=4) = \binom{6}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{6-4} = 15 \times \left(\frac{1}{625}\right) \times \left(\frac{4}{5}\right)^2 = 15 \times \frac{1}{625} \times \frac{16}{25} = \frac{15 \times 16}{15625} = \frac{240}{15625} \).

Probability for Exactly 5 Workers (P(X=5))

Here, \( n=6 \), \( k=5 \), \( p = \frac{1}{5} \), \( q = \frac{4}{5} \).

• Calculate the combination: \( \binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6}{1} = 6 \).
• Calculate the probability: \( P(X=5) = \binom{6}{5} \left(\frac{1}{5}\right)^5 \left(\frac{4}{5}\right)^{6-5} = 6 \times \left(\frac{1}{3125}\right) \times \left(\frac{4}{5}\right)^1 = 6 \times \frac{1}{3125} \times \frac{4}{5} = \frac{6 \times 4}{15625} = \frac{24}{15625} \).

Probability for Exactly 6 Workers (P(X=6))

Here, \( n=6 \), \( k=6 \), \( p = \frac{1}{5} \), \( q = \frac{4}{5} \).

• Calculate the combination: \( \binom{6}{6} = \frac{6!}{6!(6-6)!} = 1 \).
• Calculate the probability: \( P(X=6) = \binom{6}{6} \left(\frac{1}{5}\right)^6 \left(\frac{4}{5}\right)^{6-6} = 1 \times \left(\frac{1}{15625}\right) \times \left(\frac{4}{5}\right)^0 = 1 \times \frac{1}{15625} \times 1 = \frac{1}{15625} \).

Summing the Probabilities

Now, we add the probabilities calculated above:

\( P(X \ge 4) = P(X=4) + P(X=5) + P(X=6) \)

\( P(X \ge 4) = \frac{240}{15625} + \frac{24}{15625} + \frac{1}{15625} \)

\( P(X \ge 4) = \frac{240 + 24 + 1}{15625} = \frac{265}{15625} \)

Simplifying the Result

The fraction \( \frac{265}{15625} \) can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 5.

\( \frac{265 \div 5}{15625 \div 5} = \frac{53}{3125} \)

Therefore, the probability that 4 or more workers out of 6 chosen at random will suffer from the disease is \( \frac{53}{3125} \).