Question

Let the number of chocolate chips in a certain type of cookie have a Poisson distribution. We want the probability that a cookie of this type contains at least two chocolate chips to be greater than \(0.99 .\) Find the smallest value of the mean that the distribution can take.

Short answer

The smallest mean value of the distribution that ensures a probability of more than \(0.99\) for getting at least two chocolate chips is approximately \(4.61\).

Step-by-step solution

Understand Poisson probability distribution

A random variable X that has a Poisson distribution represents the number of events occurring in a fixed interval of time or space. The probability that there are exactly k events in an interval is given by the equation: \(P(X=k) = \frac{λ^k e^{-λ}}{k!}\), where \(λ\) is the mean number of occurrences, \(e\) is the base of the natural logarithms and \(k!\) is the factorial of k.

Set up the inequality

We need to find a value for \(λ\) such that a cookie containing at least two chocolate chips (2 or more) is more than \(0.99\). As it is easier to calculate probabilities of less than or equal events in a Poisson distribution, the situation can be rewritten as the probability that a cookie contains less than 2 (0 or 1 chips) is less than \(0.01\), which is the complement of \(0.99\). So, the inequality to solve becomes: \(P(X<2)= P(X=0)+P(X=1) < 0.01\)

Substitute equation and solve

Substitute the Poisson probability formula into the inequality for the probabilities that X=0 and X=1. Therefore, the inequality becomes: \(P(X=0) + P(X=1) = \frac{λ^0 e^{-λ}}{0!} + \frac{λ^1 e^{-λ}}{1!} < 0.01.\) This simplifies to \(e^{-λ}(1 + λ) < 0.01.\) Solving this inequality for \(λ\) using numerical methods, you get \(λ > 4.60517.\)