SOLUTION: Ankita enjoys her pizza. So much so that she eats pizza every night. she likes both cheese and mushroom, and each night she chooses randomly and independently between these two

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Question 1120611: Ankita enjoys her pizza. So much so that she eats pizza every night. she
likes both cheese and mushroom, and each night she chooses randomly
and independently between these
two options. Given that each night she picks the mushroom option with
probability 0.35, find the probability that she eats between 2 and 4
(inclusive) mushroom pizzas this week
(7 days).
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
probability of choosing mushroom pizza is .35
probability of choosing cheese pizza is 1 - .35 = .65

formula to use is:

p(x) = p^x * q^(n-x) * c(n,x)

p is the probability she chooses mushroom pizza on any given night.
q is the probability she chooses cheese pizza on any given night.
n is the total possible nights she gets to choose.
x is the total number of nights she chooses mushroom pizza

the probability she will choose 2 mushroom pizzas in the week gets you:

p(2) = .35^2 * .65^5 * c(7,2)

the probability she will choose 3 gets you:

p(3) = .35^3 * .65^4 * c(7,3)

the probability she will choose 4 gets you:

p(4) = .35^4 * .65^3 * c(7,4)

the probability she chooses mushroom pizza on 2 to 4 night is equal to p(2) + p(3) + p(4).

c(n,x) = n! / (x! * (n-x)!)

for example:

c(7,2) = 7! / (2! * 5!) = (7 * 6 * 5!) / (2! * 5!) = (7 * 6) / (2 * 1) = 42 / 2 = 21.

all the probabilities from 0 to 7 are shown in the following spreadsheet printout.

the total probability is 1 as it should be.

the probability of choosing mushroom pizza on 2 to 4 nights in the week is equal to p(2) + p(3) + p(4) which is equal to 0.710594 as shown in the excel printout.