SOLUTION: Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that fewer than 2 prefer brand C is
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Question 1201996: Suppose that past history shows that 60% of college students prefer Brand C cola. A sample of 5 students is to be selected. The probability that fewer than 2 prefer brand C is Answer by math_tutor2020(3816) (Show Source):
n = 5 = sample size
p = 0.6 = probability of success
x = number of people, in the sample, who prefer brand C cola
x is chosen from the set {0,1,2,3,4,5}
This is a binomial distribution problem because:
There are two outcomes. A person likes the cola or they don't.
Each trial is independent of any other. One person doesn't influence another's tastes.
The probability someone likes the cola is the same for each trial.
B(x) = binomial probability
B(x) = (n C x)*(p^x)*(1-p)^(n-x)
B(x) = (5 C x)*(0.6^x)(1-0.6)^(5-x)
B(x) = (5 C x)*(0.6^x)(0.4)^(5-x)
The nCx notation refers to the nCr formula.
It is known as the binomial coefficient.
Compute the probability of getting x = 0 people who like the cola.
B(x) = (5 C x)*(0.6^x)(0.4)^(5-x)
B(0) = (5 C 0)*(0.6^0)(0.4)^(5-0)
B(0) = 1*(0.6^0)(0.4)^5
B(0) = 0.01024
This value is exact without any rounding done to it.
Do a similar set of steps for x = 1
B(x) = (5 C x)*(0.6^x)(0.4)^(5-x)
B(1) = (5 C 1)*(0.6^1)(0.4)^(5-1)
B(1) = 5*(0.6^1)*(0.4)^4
B(1) = 0.0768
This value is exact without any rounding done to it.
The phrasing "fewer than 2" means we will add up the results of B(0) and B(1) to get the final answer.
P(fewer than 2) = B(0) + B(1)
P(fewer than 2) = 0.01024 + 0.0768
P(fewer than 2) = 0.08704
There's an 8.704% chance of getting fewer than 2 people in the sample that like brand C cola.
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