Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, 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]
Here we're given
p=0.10 [ success = defective ]
n=3
(a) x=0
[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]=C(3,0)0.1^0(1-0.1)^{3-0}[/tex]
[tex]=1(1)(0.729)[/tex]
[tex]=0.729[/tex]
(b) x=2
[tex]P(x)=C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]=C(3,2)0.1^2(1-0.1)^{3-2}[/tex]
[tex]=3(0.01)(0.9)[/tex]
[tex]=0.027[/tex]
(c) x ≥ 2
[tex]P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}[/tex]
[tex]=P(2)+P(3)[/tex]
[tex]=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}[/tex]
[tex]=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}[/tex]
[tex]=3(0.01)(0.9)+1(0.001)1[/tex]
[tex]=0.027+0.001[/tex]
[tex]=0.028[/tex]
