Questions: To reduce laboratory costs, water samples from six public swimming pools are combined for one test for the presence of bacteria. Further testing is done only if the combined sample tests positive. Based on past results, there is a 0.006 probability of finding bacteria in a public swimming area. Find the probability that a combined sample from six public swimming areas will reveal the presence of bacteria. Is the probability low enough so that further testing of the individual samples is rarely necessary? The probability of a positive test result is (Round to three decimal places as needed.)

Solution Steps

We are tasked with finding the probability that a combined sample from six public swimming areas will reveal the presence of bacteria. The probability of finding bacteria in a public swimming area is given as \( p = 0.006 \). We will use the binomial distribution to calculate the probability of at least one positive test result in six trials.

The number of trials \( n \) is 6 (the number of swimming pools), and we want to find the probability of at least one success (presence of bacteria), which can be expressed as \( P(X \geq 1) \). This can be calculated using the complement rule:

\[ P(X \geq 1) = 1 - P(X = 0) \]

Where \( P(X = 0) \) is the probability of finding no bacteria in any of the samples.

Using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

For \( x = 0 \):

\[ P(X = 0) = \binom{6}{0} \cdot (0.006)^0 \cdot (0.994)^6 \]

Calculating this gives:

\[ P(X = 0) = 1 \cdot 1 \cdot (0.994)^6 \approx 0.964 \]

Now we can find \( P(X \geq 1) \):

\[ P(X \geq 1) = 1 - P(X = 0) \approx 1 - 0.964 = 0.036 \]

The probability of a positive test result is approximately \( 0.035 \) when rounded to three decimal places.

Final Answer