- 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.
Solution:
The Poisson approximation is given as:
\[P(X = k) \approx \frac{\lambda^k}{k!}e^{-\lambda}\]
The Poisson distribution goes to infinity, therefore it is best to find the complement of the probability that a given cookie will have less than 5 raisins.
\[ \begin{align*}P(X >= 5) &= 1 - P(X < 5) \\ &= 1 - \bigg( \frac{1.2^4}{4!}e^{-1.2} \bigg) + \bigg( \frac{1.2^3}{3!}e^{-1.2} \bigg) + \bigg( \frac{1.2^2}{2!}e^{-1.2} \bigg) + \bigg( \frac{1.2^1}{1!}e^{-1.2} \bigg) + \bigg( \frac{1.2^0}{0!}e^{-1.2} \bigg) \\ &\approx 1 - 0.9922542 \\ &\approx 0.0077458 \end{align*}\]