Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=20)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (0.05)4. All trials are random and independent of the othersThe number of successes(defects), x, is then given by[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]where[tex]C(n,x)=\frac{n!}{x!(n-x)!}[/tex]
For n=20, p=0.05,
the mean number of defects is np=20*0.05=1
[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]P(X=3)=C(20,3)0.05^3(1-0.05)^{20-3}[/tex]
[tex]=1140(1.25*10^{-4}(0.41812)[/tex]
[tex]=0.05958[/tex]
Answer: probability of exactly 3 defects in a sample of 20 is 0.05958
