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 %.
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
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
Answer => .25
To calculate the probability of a customer ordering a five-topping pizza with a classic crust and extra cheese, we need to determine how many possible combinations of toppings there are and then determine the specific combination we are interested in.
First, let's consider the number of topping options available. The question states that there are 10 topping options:
- Extra cheese
- Pepperoni
- Sausage
- Onions
- Banana peppers
- Jalapenos
- Tomatoes
- Fresh herbs
- Mushrooms
- Black olives
Next, we need to determine the number of possible combinations. Since the customer can choose five toppings, we need to calculate the number of ways to choose five items from a set of ten. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!)
In this case, n = 10 (the number of topping options) and r = 5 (the number of toppings the customer can choose). Plugging in these values gives us:
C(10, 5) = 10! / (5!(10-5)!)
= 10! / (5! * 5!)
= (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1)
= 252
Therefore, there are 252 possible combinations of five toppings from the given options.
Now, we need to determine how many of these combinations include a classic crust and extra cheese. Since the crust is fixed to be classic and the extra cheese is one of the topping options, there is only one way to include both of them.
Therefore, the probability of a customer ordering a five-topping pizza with a classic crust and extra cheese is:
1 / 252 ≈ 0.3968
Converting this to a percentage, we get:
0.3968 * 100% ≈ 39.68%
So, the probability is approximately 39.68%.