how to fit a binomial distribution of screw which has defective
Answers
Answer:
A machine manufacturing screws is known to produce 5% defective screws. In a random sample of 15 screws, what is the probability that there are exactly three defective screws?
The answer to the problem:
P(d= 3) = 15C3 * 0.05^3 * 0.95^12 = 455 * 0.000000125 * 0.540360 = 0.030733
0.030733 or app. 3%
The understanding of the answer:
We want exactly three screws to be defective. Each of these screws have only 5% probability of being defective. So If we had only three screws, and we want all three to be defective, the probability is 0.05^3
But we do not have only 3 screws. We have 15 screws, and therefore we want the balance 12 to be defect free. So for these 12 screws, the probability that all of them are defect free is 0.95 ^ 12
In arriving at the above, we have split the 15 screws into two groups of 3 & 12. In how many ways can we select 3 out of 15, leading to a group of 3 and another group of balance 12. It is given by the formula for number of combinations for 3 out of 15, (15*14*13) / (1*2*3), or in short form 15C3.
Therefore the final answer we want is the multiplication of all three of above, which is the formula written in the beginning of this answer.
The result we obtained can be got from tables of Binomial distribution. The name Binomial distribution comes from Binomial expansion. Look up the binomial expansion of (p+q)^n. Each of the terms in this expansion is a member of the binomial distribution. Specifically this is the 4th term in (0.95+0.05)^15. The full expansion of this will be
I hope this answer helpful for you