SOLUTION: Suppose that you and two friends go to a restaurant, which last month filled approximately 73% of the orders correctly.
What is the probability that at least two of the three
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What is the probability that at least two of the three
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Question 1190727: Suppose that you and two friends go to a restaurant, which last month filled approximately 73% of the orders correctly.
What is the probability that at least two of the three orders will be filled correctly? Answer by math_tutor2020(3817) (Show Source):
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This is a binomial probability distribution problem.
We have two outcomes: order filled correctly vs not filled correctly
Each trial has the same probability p = 0.73 (the order being filled correctly)
Each trial is independent of one another.
The sample size is n = 3
The binomial formula is
B(k) = (nCk)*(p)^k*(1-p)^(n-k)
B(k) = (3Ck)*(0.73)^k*(1-0.73)^(3-k)
B(k) = (3Ck)*(0.73)^k*(0.27)^(3-k)
where the nCk refers to the nCr combination formula.
Now plug in k = 2 to find the probability of exactly 2 orders filled correctly
B(k) = (3Ck)*(0.73)^k*(0.27)^(3-k)
B(2) = (3C2)*(0.73)^2*(0.27)^(3-2)
B(2) = (3)*(0.73)^2*(0.27)^(1)
B(2) = 0.431649
Repeat for k = 3
B(k) = (3Ck)*(0.73)^k*(0.27)^(3-k)
B(3) = (3C3)*(0.73)^3*(0.27)^(3-3)
B(3) = (1)*(0.73)^3*(0.27)^(0)
B(3) = 0.389017
Add up the results to find the probability of at least 2 (aka 2 or more) orders filled correctly.
0.431649 + 0.389017 = 0.820666
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