Again the crust will not change the probability.
Here the customer chooses 5 toppings out of 10 in C(10,5)=10!/(5!5!)=252 ways. This is the sample space.
If the customer is to have onions, jalapenos and tomatoes, he has two other choices out of 7 remaining toppings in C(7,2)=7!/(2!5!)=21 ways.
The probability is therefore:
21/252=1/12
A pizza place sells five topping pizzas with the following options. Crust: Classic, Crispy Toppings: Extra cheese, pepperoni, sausage, onions, banana peppers, jalapenos, tomatoes, fresh herbs, mushrooms, black olives.
What is the probability that a customer will order a five topping pizza including onions, jalapenos, and tomatoes?
Explain how to do it in word please?
1 answer