Given : Sample size : n= 5
The proportion nonconforming : p= 0.10
Binomial probability formula :-
[tex]P(x)=^nC_x p^{x}(1-p)^{n-x}[/tex]
The probability of zero nonconforming unit in the sample :-
[tex]P(0)=^5C_0 (0.10)^{0}(1-0.1)^{5}\\\\=(1)(0.9)^5\ \ [ \because\ ^nC_0=1]=0.59049[/tex]
∴ The probability of zero nonconforming unit in the sample= 0.59049
The probability of one nonconforming unit in the sample :-
[tex]P(1)=^5C_1 (0.10)^{1}(0.9)^{4}\\\\=(5)(0.1)(0.9)^5\ \ [ \because\ ^nC_1=n]=0.295245[/tex]
∴ The probability of one nonconforming unit in the sample=0.295245
The probability of 2 or more nonconforming units in the sample :-
[tex]P(X\geq2)=1-(P(0)+P(1))=1-(0.59049+0.295245)\\\\=1-0.885735=0.114265[/tex]
∴ The probability of 2 or more nonconforming units in the sample=0.114265
