Question

at a certain pizzeria, it is known that 12% of orders are for extra - large pizzas. what is the expected number of pizzas that will be ordered until the first extra - large pizza is ordered?
6 pizzas
8 pizzas
8.33 pizzas
12 pizzas

Explanation:

Step1: <Identify the distribution>

This is a geometric distribution problem. In a geometric distribution, the probability of success (here, success is an extra - large pizza order) is \(p\). The formula for the expected value \(E(X)\) of a geometric distribution is \(E(X)=\frac{1}{p}\).

Step2: <Substitute the value of \(p\)>

Given that \(p = 0.12\) (since \(12\%=0.12\)). Then \(E(X)=\frac{1}{0.12}=\frac{100}{12}=\frac{25}{3}\approx8.33\)

Answer:

8.33 pizzas