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 0.004 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.)

Given: - The probability of finding bacteria in a public swimming area is 0.004. - The combined sample is from six public swimming areas. Let's denote: - \( P(\text{Bacteria}) \) as the probability of finding bacteria in a public swimming area. - \( P(\text{Positive Test}) \) as the probability of a positive test result for the combined sample. The probability of finding bacteria in a public swimming area is 0.004, so: \[ P(\text{Bacteria}) = 0.004 \] The probability of a positive test result for the combined sample can be calculated using the binomial probability formula: \[ P(\text{Positive Test}) = \sum_{k=1}^{6} \binom{6}{k} (0.004)^k (1-0.004)^{6-k} \] Let's calculate the probability of a positive test result for the combined sample. Find the sum by following steps: - step0: Solution: \(\sum _{k=1}^{6} { }_{6}C_{k}\times 0.004^{k}\left(1-0.004\right)^{6-k}\) - step1: Subtract the numbers: \(\sum _{k=1}^{6} { }_{6}C_{k}\times 0.004^{k}\times 0.996^{6-k}\) - step2: Expand the expression: \(\sum _{k=1}^{6}\frac{6!}{k!\times \left(6-k\right)!}\times 0.004^{k}\times 0.996^{6-k}\) - step3: Calculate the value: \(\sum _{k=1}^{6}\frac{720}{k!\times \left(6-k\right)!}\times 0.004^{k}\times 0.996^{6-k}\) - step4: Convert the expressions: \(\sum _{k=1}^{6}\frac{720}{k!\times \left(6-k\right)!}\times \left(\frac{1}{250}\right)^{k}\times 0.996^{6-k}\) - step5: Convert the expressions: \(\sum _{k=1}^{6}\frac{720}{k!\times \left(6-k\right)!}\times \left(\frac{1}{250}\right)^{k}\left(\frac{249}{250}\right)^{6-k}\) - step6: Multiply the terms: \(\sum _{k=1}^{6}\frac{9\left(\frac{249}{250}\right)^{6-k}}{k!\times \left(6-k\right)!\times 2^{k-4}\times 5^{3k-1}}\) - step7: Rewrite the expression: \(\sum _{k=1}^{6}\frac{9\left(\frac{249}{250}\right)^{6}\left(\frac{249}{250}\right)^{-k}}{k!\times \left(6-k\right)!\times \frac{1}{16}\times 2^{k}\times \frac{1}{5}\times 5^{3k}}\) - step8: Rewrite the expression: \(\sum _{k=1}^{6}\frac{1}{3051757812500}\left(k!\right)^{-1}\left(\left(6-k\right)!\right)^{-1}\times 250^{-k}\left(\frac{249}{250}\right)^{-k}\times 9\times 249^{6}\) - step9: Convert the expressions: \(\frac{1}{3051757812500}\left(1!\right)^{-1}\left(\left(6-1\right)!\right)^{-1}\times 250^{-1}\left(\frac{249}{250}\right)^{-1}\times 9\times 249^{6}+\frac{1}{3051757812500}\left(2!\right)^{-1}\left(\left(6-2\right)!\right)^{-1}\times 250^{-2}\left(\frac{249}{250}\right)^{-2}\times 9\times 249^{6}+\frac{1}{3051757812500}\left(3!\right)^{-1}\left(\left(6-3\right)!\right)^{-1}\times 250^{-3}\left(\frac{249}{250}\right)^{-3}\times 9\times 249^{6}+\frac{1}{3051757812500}\left(4!\right)^{-1}\left(\left(6-4\right)!\right)^{-1}\times 250^{-4}\left(\frac{249}{250}\right)^{-4}\times 9\times 249^{6}+\frac{1}{3051757812500}\left(5!\right)^{-1}\left(\left(6-5\right)!\right)^{-1}\times 250^{-5}\left(\frac{249}{250}\right)^{-5}\times 9\times 249^{6}+\frac{1}{3051757812500}\left(6!\right)^{-1}\left(\left(6-6\right)!\right)^{-1}\times 250^{-6}\left(\frac{249}{250}\right)^{-6}\times 9\times 249^{6}\) - step10: Calculate: \(\frac{729\times 83^{5}}{122070312500000}+\frac{243\times 83^{4}}{48828125000000}+\frac{27\times 83^{3}}{12207031250000}+\frac{186003}{48828125000000}+\frac{747}{122070312500000}+\frac{1}{244140625000000}\) - step11: Calculate: \(\frac{5801092813999}{244140625000000}\) The probability of a positive test result for the combined sample from six public swimming areas is approximately 0.023761. This probability is low enough that further testing of the individual samples is rarely necessary.