If a product with a probability of defect of 0.1 is a random 8, then the number of defects is a random quantity and find the law of its distribution.

Question

help me to solve this

Answer 1

My interpretation of this question is that a particular product has a probability of 0.1 of being defective. In a random sample of 8 products devise a law of probability distribution for the number of defective products in the sample.

A product is either defective or it isn’t, so that’s a two-state, or binomial, scenario. If we write p as the probability that the product is defective, and q that it is not defective then q=1-p. We can then use the binomial expansion to determine the probability of d defects in the sample for a given d.

(p+q)ⁿ=pⁿ+np^n-1q+n(n-1)p^n-2q²/2!+…+ⁿC_dp^n-dq^d+…n(n-1)p²q^n-2/2!+npq^n-1+qⁿ, where n=8 (sample size), p=0.1 and q=0.9.

ⁿC_d=n!/(d!(n-d)!) is the combination function which gives the number of ways of combining d items out of n items.

The function for the probability of exactly d defective products is P(d)=ⁿC_dp^n-dq_d=8C_d×0.1^d×0.9^8-d.

For example, when d=2, P(2)=28×0.1²×0.9⁶=0.1488 or 14.88%, the probability that 2 products in a random sample of 8 are defective.