SOLUTION: A survey finds customers overpay for 4 out of every 10 items. Suppose a customer purchases 15 items.
a) the probability the customer overpays on none of the items is______? (rou
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Question 761278: A survey finds customers overpay for 4 out of every 10 items. Suppose a customer purchases 15 items.
a) the probability the customer overpays on none of the items is______? (round decimal to the nearest thousandth).
b) The probability the customer overpays on at least 3 items is______? (round decimal to the nearest thousandth).
-I got .104 for the first answer is that correct? I can't figure out the second one!
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A survey finds customers overpay for 4 out of every 10 items. Suppose a Cheers,
Stan H.
customer purchases 15 items.
----
Binomial Problem with n = 15 and p(overpay) = 0.4
--------------------------------------------------------
a) the probability the customer overpays on none of the items is 0.6^15 =
0.000470.. (round decimal to the nearest thousandth).
--------------------------------------------------------
b) The probability the customer overpays on at least 3 items is
P(3<= x <=15) = 1 - binomcdf(15,0.4,2) = 0.9729
(round decimal to the nearest thousandth).
------------------------------------------------------
Cheers,
Stan H.
================
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
No.
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 !})
For part a you need \ =\ {{15}\choose{0}}\left(0.4\right)^0\left(0.6\right)^{15})
I'll let you do your own arithmetic. (hint: n choose 0 = 1)
For part b, the straight forward approach is to calculate the above for 
, then 
, then 5, 6, ... and so on up to 15. And then you have to add up all of your answers for these results. Lots of VERY nasty arithmetic. Looks like this:
\ =\ \sum_{i\,=\,k}^n\,{{n}\choose{i}}\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i})
Better approach: The probability of AT LEAST 3 is equal to 1 minus the probability of LESS THAN 3. The probability of less than 3 is the probability of zero (which is what you just calculated in part a) plus the probability of 1 plus the probability of 2. Don't forget to subtract the sum from 1.
\ =\ 1\ -\ \sum_{i\,=\,0}^{k\,-\,1}\,{{n}\choose{i}}\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i})
John
My calculator said it, I believe it, that settles it
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