Let the number of chocolate chips in a cookie be modeled by

Let the number of chocolate chips in a cookie be modeled by a Poisson distribution.
a. If the average number of chocolate chips per cookie is 5, what is the probability that a randomly selected
cookie contains at least two chocolate chips?
b. If I take three cookies, what is the probability that at least two of my cookies have at least two chocolate
chips?
c. Suppose that the cookie company wants to evaluate the amount of chocolate chips they put in their
cookies. They want to have at least a 99% chance that a randomly chosen cookie will contain at least two
chocolate chips. To ensure this, what is the smallest value that the mean can be? (Find this value to one
decimal place)

Solution

a)

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 =    5      
          
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   ) =    0.040427682
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.959572318

b)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    3      
p = the probability of a success =    0.959572318      
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   ) =    0.004771043
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.995228957 [ANSWER]

c)

As

P(at least 2) = 1 - P(0) - P(1)

As, in Poisson distribution,

P(0) = e^-u
P(1) = u e^-u

Then

P(at least 2) = 1 - e^-u - u e^-u = 0.99

Solving for u,

u = 6.7 [answer]