An order for 1000 parts is received The probability that an

An order for 1000 parts is received. The probability that an individual part is defective is 10^-3. Let the random variable X(zeta) = #defective parts in this order of 1000 parts. Assuming each part is defective independently of any other, model this as Bernoulli trials and compute the probability that we have no defective parts in the order. Approximate your answer in part a using the Poisson PMF and compare your answers. Using the approximation in part b. what is the probability of more than 2 defective parts in this order?

Solution

1a)

P(no defective) = (1-0.001)^1000 = 0.367695425 [ANSWER]

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1b)

Here, the mean of the Poisson distrbution is

mean = n p = 1000*0.001 = 1.

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 =    1      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.367879441 [ANSWER]

which is very close to that in part a).

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1c)

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    1      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   2   ) =    0.919698603
          
Thus, the probability of at least   3   successes is  
          
P(more than   2   ) =    0.080301397 [answer]