Question 916981
this is a binomial probability.
the formula is:
p(x) = c(n,x) * p^x * q^(n-x)


x is the number of defectives items.
n is the total number of items.
p is the probability of getting one that is defective.
q is the probability of getting one that is not defective
the probability of getting one that is not defective is equal to 1 minus the probability of getting one that is defective.


in this problem:
n = 17
x = 0 to 6
p = 6/17 = .352941...
q = 11/17 = .647058...


c(n,x) is the combination of n things taken x at a time where order doesn't matter.


this formula is:
c(n,x) = n! / (x! * (n-x)!)


you want to find the probability that more than 4 are defective out of 6 selected.


this means the probability that 5 or 6 are defective.


4 defective passes the test and the box is no rejected.
5 or 6 defective fails the test and the box is rejected.


p(5) = c(6,5) * p^5 * q^1 = 6 * (6/17)^5 * (11/17)^1 = .021262...
p(6) = c(6,6) * p^6 * q^0 = 1 * (6/17)^6 * (11/17)^0 = .001932...


add these together and you get p(5 or 6) = .023195 rounded to 6 decimal places.


that's your solution.


the probability that the box will be rejected = .023195 *****


the following excel spreadsheet shows all the probabilities.


<img src = "http://theo.x10hosting.com/2014/102601.jpg" alt="$$$" </>


you can see that the total probability is equal to 1.
you can slso see that the probability of 5 or 6 defectives when you pick 6 out of the box of 17 is equal to .023195.
the column marked x contains the number of defectivdes.
the column markes p(x) contains the probability of getting that many defectives.