Re: probability question
This can be solved by a binomial consideration.
The whole process consists of 1000 bernoulli tests, each of which amounts to pick up a screw and check whether it's good or bad. Let's assume that the probability for a good screw is p while for a bad screw q. p+q=1. We'll get back to them shortly.
Set n=1000, and m=300, where m=300 comes from the number 70% which means 30% of the screws (=300) can be bad. Therefore, the probability that 70% of the screws are good is the following
Sum[C(n,i)q^i * p^(n-i),{i=0, to m}],
where I have used the notations as in Mathematica. C(n,i)=n!/(i!(n-i)!), the sum is performed for i from 0 to m (=300).
Ok, let's get to the hard part, i.e., to determine p and q.
Set A1 = the screw is longer than 0.9cm, B1 = the screw is shorter than 0.9cm. Since the length is uniformly distributed and you can easily find out that 0.9 is right at the middle of (0.65, 1.15), P(A1)=1/2 and P(B1)=1/2.
Set A2 = screw has a head, B2= screw doesn't have a head. According to the condition, P(A2)=49/50 and P(B2)=1/50.
Since a screw is bad iff the screw belongs to B1UB2 (the union of B1 and B2), we have
q=P(B1UB2)=1-P(A1A2)=1-P(A1|A2)P(A2)=1-(1/2)*(49/50),
where I have used the relation B1UB2 = {whole sample space} - A1A2, A1 is independent of A2 ( I supposed the head and the shank are separate parts of a screw).
Therefore, p=1-q=(1/2)*(49/50).
Sorry, busy lately and don't have time to answer more questions.