NOTE: read through the interpretations of the question before deciding if the answer is applicable to you.
A classic crust is one choice out of three in the first step of the experiment. The probability for step 1 is therefore 1/3.
"Extra cheese, pepperoni, sausage, onions, banana peppers, jalapenos, tomatoes, fresh herbs, mushrooms, black olives "
gives a count of 10 toppings, assuming "banana peppers" is a single topping for the lack of comma.
Given that the customer chooses extra cheese, he has a choice of 4 more toppings out of the remaining 9 (assuming no repeated toppings allowed).
Number of choices of the remaining toppings is therefore
C(9,4)=9!/(5!4!)=126
Total number of choices of toppings is 2^10 (by choosing any combination, including nothing to everything).
So probability of choosing 5 toppings including extra cheese is 126/1024
Overall probability is the product of probabilities of the two steps, namely
P(classic,extra cheese+4 others)
=(1/3)(126/1024)
=21/512
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 with a classic crust and extra cheese?
Give answer as %.
3 answers
Sorry I have not noticed that the pizza place sells 5-topping pizzas. The answer (in probability terms) will therefore be different.
See other post for details:
http://www.jiskha.com/display.cgi?id=1337879558
See other post for details:
http://www.jiskha.com/display.cgi?id=1337879558
Answer => .25