SOLUTION: Historically, 9 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database.
a. What is the probability
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Question 596785: Historically, 9 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database.
a. What is the probability that none of the next 16 repeat customers who call will have an incorrect address?
b. What is the probability that five customer who call will have an incorrect address?
c. What is the probability that six customers who call will have an incorrect address?
d. What is the probability that fewer than seven customers who call will have an incorrect address?
All answers should be rounded to four decimal places.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
a.
so
\ =\ {{16}\choose{0}} \left(0.09\right)^0\left(0.91\right)^{16}\ =\ (0.91)^{16})
because 
and 
b. Presuming you mean exactly 5 out of 16:
so
\ =\ {{16}\choose{5}} \left(0.09\right)^5\left(0.91\right)^{11})
c. Again presuming 6 out of 16:
so
\ =\ {{16}\choose{6}} \left(0.09\right)^6\left(0.91\right)^{10})
d. Again presuming less than 7 out of 16:
so
\ =\ \sum_{i\,=\,0}^6\,{{16}\choose{i}} \left(0.09\right)^i\left(0.91\right)^{16\,-\,i})
Hint: You have already done three terms of this seven-term sum, namely i = 0 (part a), i = 5 (part b), and i = 6 (part c).
You have a LOT of arithmetic to do. Better get busy.
John
My calculator said it, I believe it, that settles it
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