Question 1104764
this is a binomial probability type problem.


the probability that the process will produce a defective part is .2.


this makes p = .2


q is the probability that the process will not produce a defective part.


the probability of that is equal to 1 - .2 = .8


this makes q = .8


the formula for binomial probability is p(x) = p^x * q^(n-x) * c(n,x)


c(n,x) is equal to n! / (x! * (n-x)!)


it's the number of ways you can get sets of x out of a set of n when the order of the elements within the set is not important.


so, your formula is p(x) = p^x * q^(n-x) = c(n,x).


n is equal to 12.


when x is equal to 6, the formula becomes p(6) = (.2)^2 * (.8)^6 * c(12,6).


when x is equal to 7, the formula becomes p(7) = (.2)^7 * (.8)^5 * c(12,7).


evaluate these equations and you get:


p(6) = .0155021476


p(7) = .0033218888


add them together and the probability of getting 6 or 7 defective parts is equal to .0188240364.


your solution is that the probability of getting 6 or 7 defective parts out of a batch of 12 parts is equal to .0188240364.


the following excel spreadsheet printout shows all the probabilities and confirms that the sum of all the probabilities must be equal to 1 and also confirms that the answer is correct.


<img src= "http://theo.x10hosting.com/2017/121404.jpg" alt="$$$">