Answer:
The probability of a positive test result is 0.017919
Option C is correct.
The probability is quite low, indicating that further testing of the individual samples will be a rarely necessary event.
Step-by-step explanation:
Probability of finding bacteria in one public swimming pool = 0.009
We now require the probability of finding bacteria in the combined test of two swimming pools. This probability is a sum of probabilities.
Let the two public swimming pools be A and B respectively.
- 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!!!
