To reduce laboratory costs, water samples from two 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 0.009 probability of finding bacteria in a public swimming area. Find the probability that a combined sample from twotwo 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 _____.

Is the probability low enough so that further testing of the individual samples is rarely necessary?
A. The probability is quite low, indicating that further testing is necessary for all of the combined samples.
B. The probability is quite low, indicating that further testing is not necessary for any of the combined samples.
C. The probability is quite low, indicating that further testing of the individual samples will be a rarely necessary event.
D. The probability is quite low, indicating that further testing of the individual samples will not be a rarely necessary event.

- It is possible for public swimming pool A to have bacteria and public swimming pool B not to have bacteria. We would obtain a positive result from testing a mixed sample of both public swimming pools.

- It is also possible for public swimming pool A to not have bacteria and public swimming pool B to have bacteria. We would also obtain a positive result from testing a mixed sample of both public swimming pools.

- And lastly, it is possible that both swimming pools both have bacteria in them. We will definitely get a positive result from this too.

So, if P(A) is the probability of the event of bacteria existing in public swimming pool A

And P(B) is the probability of the event of bacteria existing in public swimming pool B

P(A') and P(B') represent the probabilities of bacteria being absent in public swimming pool A and public swimming pool B respectively.

P(A) = P(B) = 0.009

P(A') = P(B') = 1 - 0.009 = 0.991

Since the probabilities for each public swimming pool is independent of the other.

P(A or B) = P(A n B') + P(A' n B) + P(A n B)

= P(A)×P(B') + P(A')×P(B) + P(A)×P(B)

= (0.009×0.991) + (0.991×0.009) + (0.009×0.009)

= 0.008919 + 0.008919 + 0.000081

= 0.017919

Evidently, a probability of 0.017919 (1.7919%) indicates an event with a very low likelihood. A positive result is expected only 1.7919% of the time.

Hence, we can conclude that the probability is quite low, indicating that further testing of the individual samples will be a rarely necessary event.

Hope this Helps!!!