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+npn-1q+n(n-1)pn-2q2/2!+…+nCdpn-dqd+…n(n-1)p2qn-2/2!+npqn-1+qn, where n=8 (sample size), p=0.1 and q=0.9.
nCd=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)=nCdpn-dqd=8Cd×0.1d×0.98-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.