Understanding the Probability Problem: At Least One Worker Suffering

The question asks for the probability that at least one out of 6 workers suffers from a disease, given the incidence rate for a single worker. We are told that the chance of suffering from the disease for an individual worker is \(33\frac{1}{3}\%\).

First, let's convert the given percentage into a simple 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 \(p = \frac{1}{3}\).

The number of workers is \(n = 6\). We want to find the probability that at least one of these 6 workers suffers from the disease.

Calculating the Probability of At Least One Event

When dealing with "at least one" probability questions, it is often easier to calculate the probability of the complementary event. The complementary event to "at least one worker suffers" is "no worker suffers".

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

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

Assuming that the suffering of each worker is an independent event, the probability that none of the 6 workers suffer from the disease is the product of the probabilities that each individual worker does not suffer.

Probability (none suffer) = \(P(\text{worker 1 does not suffer and worker 2 does not suffer ... and worker 6 does not suffer})\)

Since they are independent events, this is:

\(P(\text{none suffer}) = q \times q \times q \times q \times q \times q = q^6\)

Substitute the value of \(q\):

\(P(\text{none suffer}) = \left(\frac{2}{3}\right)^6\)

Let's calculate the numerator and denominator separately:

• Numerator: \(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\)
• Denominator: \(3^6 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 9 = 81 \times 9 = 729\)

So, the probability that none of the 6 workers suffer is:

\(P(\text{none suffer}) = \frac{64}{729}\)

Now, we can find the probability that at least one worker suffers using the complement rule:

\(P(\text{at least one suffers}) = 1 - P(\text{none suffer})\)

\(P(\text{at least one suffers}) = 1 - \frac{64}{729}\)

To subtract the fraction from 1, we can write 1 as \(\frac{729}{729}\):

\(P(\text{at least one suffers}) = \frac{729}{729} - \frac{64}{729} = \frac{729 - 64}{729}\)

Perform the subtraction in the numerator:

\(729 - 64 = 665\)

Therefore, the probability that at least one out of 6 workers suffers from the disease is:

\(P(\text{at least one suffers}) = \frac{665}{729}\)

Summary of Probability Calculation Steps

The calculated probability is \(\frac{665}{729}\), which corresponds to one of the given options.

Revision Table: Key Probability Concepts

Additional Information on Probability Calculations

Probability is a fundamental concept in mathematics that quantifies the likelihood or chance of an event occurring. It is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.

In this problem, we used the concept of independent trials, where the outcome for one worker does not influence the outcome for another worker. When dealing with a fixed number of independent trials (like 6 workers) where each trial has only two possible outcomes (suffers or does not suffer) with fixed probabilities, this scenario fits the description of a Bernoulli process, which can be analyzed using binomial probability.

The binomial probability formula for getting exactly \(k\) successes in \(n\) trials is \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\).

In our case, "suffering" is a "success" (even though it's a negative outcome in reality, in binomial terms, it's the event we are counting).

"At least one suffers" means either 1 suffers, or 2 suffer, ..., or 6 suffer. Calculating the probability for each of these cases (P(X=1) + P(X=2) + ... + P(X=6)) would be much more complex than using the complement approach (1 - P(X=0)).

Calculating P(X=0) using the binomial formula:

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

We know that \(\binom{6}{0} = 1\) and \(\left(\frac{1}{3}\right)^0 = 1\).

So, \(P(X=0) = 1 \times 1 \times \left(\frac{2}{3}\right)^6 = \left(\frac{2}{3}\right)^6 = \frac{64}{729}\).

This confirms our earlier calculation for the probability that none suffer. The complement approach is a powerful shortcut for "at least one" problems in probability.