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Algebra 2:)

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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 %.

  • Algebra 2:) - ,

    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

  • Algebra 2 - ,

    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

  • Algebra 2:) - ,

    Answer => .25

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