Please help me with a probability task

probability question

I have an exercise and I am clueless, can you help me ?

1000 screws are daily produced in a line of production, the shanks have lengths from 0.65 cm to 1,15 cm uniformly distributed. The head always has 0,1 cm, but there are a chance that 1 among 50 screws is headless. At the end of the day a quality control eliminates:

- Screws headless
-Screws with total length less than 0.9 cm.

Determine the probability that 70% of screws are accepted by the quality control at the end of the day.

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.

probability question

Thanks!

From here it is possible to finish the exercice using o D'Moivre Laplace. Just a remark, I think that in determining the probabilities p and q the length of the head must be considered since is the total length of the screw which is considered for acceptance, so there are a slight changing in the probabilities results, but it is ok, what really matters is the overall idea.

Re: probability question

Sorry for the late response (too busy).

Yes, you are right. Here is the right place to apply that theorem, but you also take the sum directly using a program.