Assume that we are making raisin cookies. We put a box of 600 raisins into our dough mix, mix up the dough, then make from the dough 500 cookies. We then ask for the probability that a randomly chosen cookie will have 0, 1, 2, . . . raisins. Consider the cookies as trials in an experiment, and let X be the random variable which gives the number of raisins in a given cookie. Then we can regard the number of raisins in a cookie as the result of n = 600 independent trials with probability p = 1/500 for success on each trial. Since n is large and p is small, we can use the Poisson approximation with \(\lambda\) = 600(1/500) = 1.2. Determine the probability that a given cookie will have at least five raisins.
\[P(X = k) \approx \frac{\lambda^k}{k!}e^{-\lambda}\]
We know that \(\lambda\) is 1.2
Probability that a given cookie will have at least 5 raisins
\[P(X >= 5) = 1 - P(X < 5)\]
\[1 - \bigg( \frac{1.2^4}{4!}e^{-1.2} \bigg) + \bigg( \frac{1.2^3}{3!}e^{-1.2} \bigg) + \bigg( \frac{1.2^2}{2!}e^{-1.2} \bigg) + \bigg( \frac{1.2^1}{1!}e^{-1.2} \bigg) + \bigg( \frac{1.2^0}{0!}e^{-1.2} \bigg)\]
# Upper tail
ppois(5, lambda=1.2, lower=FALSE)
## [1] 0.001500225