VI The number of bacteria in a 10ml sample of swimming pool

VI: The number of bacteria in a 10ml sample of swimming pool water, Y has a Poisson distribution with mean = 1.2.

1. Supposethata10mlsampleofpoolwaterisdeclaredcontaminatedifatleasttwobacteria are found. What is the probability of that happening?

2. Suppose that we have two independent and non-overlapping 10ml samples of pool water and we mix them together. What is the probability that we find exactly 4 bacteria in this 20 ml of water?

Solution

1.

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    1.2      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.662627266
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.337372734 [ANSWER]

*********************

2.

Now that we have two 10 mL samples, the new mean is 2*1.2 = 2.4.

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    2.4      
          
x = the number of successes =    4      
          
Thus, the probability is          
          
P (    4   ) =    0.125408499 [ANSWER]