Question Thank you for your reply that you don't answer homework questions. However, I'm 35 years old and looking to broaden my expertise into lean / Six Sigma. One of the subjects is probability which I've not been so good at. I have found some questions relating to probability but cannot find the answers. I understand the rolling of dice and playing cards but applying it to other situations is causing me to fall down. If you could help with the question that would be great, even if it is put into another situation. Really looking for some guidance on the theory and the need to understand how the answer is achieved, rather than just the answer.
'Circuit boards are produced with a failure rate of 85%.
If a sample of 12 boards are taken, what is the probability that at least 2 boards are defective?'
My initial thoughts would be to multiple the probabilities together as the failure of the boards are independent, this would be: 0.85*0.85=0.7225 (72.25%)
However, I'm not sure where the sample of 12 comes in.
Answer nathan -
you've got a bit of a row to hoe if you want to become better acquainted with
6 sigma and related aspects of quality control. i can tell you how to answer
the question you asked - but you'll need to be able to do some reading on your
own to absorb some basic concepts of probability - and then go on from there.
your question involves what is called the binomial distribution - which applies
to situations where one has repeated (independent) trials - like repeatedly tossing
a biased coin and asking for the probability that x (say) number of heads will appear
in the n (say) trials. if X represents the (random) number of heads that will appear,
the probability that X=x is given by
P(X=x) = C(n,x)p^x (1-p)^(n-x).
here C(n,x) means 'n choose x' and p is the 'success probability', the chance of a head
appearing on any given trial.
in your case, n=12 and p=.85.
so the brute force answer is to compute P(X=x) for x = 2,3,...,12 and add up those
numbers, to get P(X is at least 2).
a somewhat less tedious method is to note that P(X is at least 2) = 1-P(X<1).
here (X<1) is the complementary event. to compute its probability, one need only
compute P(X=x) for x = 0 and 1.
as P(X=0) = (1-p)^12 = .15^12 and P(X=1) = 12p(1-p)^11 = 12x(.85)x(.15)^11, the answer
can be easily obtained with a scientific calculator. one can also obtain it more
easily using excel, which has among its statistical functions one for the binomial
distribution.
good luck with your endeavor to learn more about 6 sigma.
