SOLUTION: Eighty percent of married couples paid for their honeymoon themselves. You randomly select 20 married couples and ask if they paid for their honeymoon themselves. Find the probab

Question 459280: Eighty percent of married couples paid for their honeymoon themselves. You randomly select 20 married couples and ask if they paid for their honeymoon themselves. Find the probability that the number of couples who say they paid for their honeymoon themselves is:
a) exactly fifteen
b) at least twelve
c) less than twelve
Answer by math-vortex(648) (Show Source): You can put this solution on YOUR website!
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problem a. Find P(exactly 15)
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This is a binomial probability problem:
Success = couple paid for their own honeymoon.
Failure = couple did not pay for their honeymoon.
n = the number of trials = 20
x = number of successes = 15
p = probability of success = 0.8
q = 1 - p = probability of failure = 1 - 0.8 = 0.2
nCx = 20C15 = the number of ways to combine 20 objects in groups of 15
Apply the binomial formula:
b(x; n, p) = nCx * p^x * q^(n-x)
b(15; 20, 0.8) = [20C15]=(15504)(0.03518)(0.00032)=0.17456
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problem b. Find P(x>11) "probability of at least 12 successes""
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This is a cumulative binomial probability problem. You approach it the same way as problem a, but you will find multiple probabilities and add them together.
Notice that "at least 12" is the same thing as "12, 13, 14, 15, 16, 17, 18, 19, 20," so
P(at least 12) = P(x=12) + P(x=13) + P(x=14) + P(x=15) + P(x=16) + P(x=17) + P(x=18) + P(x=19) +P(x=20)
Set up each of the probabilities as you did in problem a. For example,
P(x=12) = [20C12]
(I'll leave the expansion of each term to you.)
P(x>11) = 0.02216 + 0.05454 + 0.10910 + 0.17456 + 0.21820 + 0.20536 + 0.13691 + 0.05765 + 0.01153 = 0.99001
Here is a website that will quickly do the arithmetic for you:
http://stattrek.com/Tables/Binomial.aspx
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problem c. Find P(less than 12 successes)
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The good news is that you do not need to do a bunch of binomial expansions and arithmetic to solve this problem. Since we know the probability of at least 12 successes from problem b, and we know that the sum of all the probabilities is 1.00, this becomes a simple subtraction problem.
P(less than 12 successes) = 1 - P(at least 12 success)
P(x<12) = 1 - 0.9901 = 0.0099
Hope this helps.
Good luck!

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