- A manufacturer buys bolts from two suppliers, with 40% from S1 and 60% from S2 - 1.5% of bolts from S1 are defective, and 0.5% from S2 are defective - The question asks for the expected number of bolts that must be inspected from S1 until a non-defective bolt is found

1. A manufacturer buys threaded bolts in large quantities from two suppliers S1 and S2. The bolts
are kept in a bin from where they are drawn. 40% of the bolts are from S1 and 60% from S2.
1.5% of S1 bolts are bad and .5% of bolts from S2 are bad. If bolts from S1 are inspected until
one bolt that is NOT bad is found, what is the expected number of bolts that must be inspected?
My thoughts is this is the mean for a binomial and it would be n*p? Not sure of how to actually
work this one out.
Solution
This is a geometric distribution. You are sampling until a bolt is not bad. If you
would like to derive the distribution, let p be the probability that the first bolt is good. Trivially,
then, p(1) = p. To determine p(n), if n is the first time the bolt is not bad, the first n-1 bolts had to
be bad. The probability that the first n-1 bolts are bad = (1-p) ^(n-1). The probability that the
n'th bolt is good is p. Thus, p(n) = (1-p) ^(n-1) * p. Note that this formula also works for n = 1.
The mean of a geometric distribution with probability p is 1/p. 1/(1-.015) = 1.01522842639594 .
We might want to express this as 1.0152. See
http://en.wikipedia.org/wiki/Geometric_distribution for an explanation. (We are using the first
definition.)