SOLUTION: The probability that a certain production process will produce a defective part is 0.20. If a box contains 12 parts, the probability that there will be 6 or 7 defective parts is ?
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Question 1104764: The probability that a certain production process will produce a defective part is 0.20. If a box contains 12 parts, the probability that there will be 6 or 7 defective parts is ? Answer by Theo(13342) (Show Source):
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.
