Binomial probability formula to find the probability of getting success in x trials is given by :-
[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], where n is the sample size , p is the proportion of getting success in each trial. (1)
Given : A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective.
i.e. n= 2
a) If three of the six systems are actually defective.
[tex]\Rightarrow\ p=\dfrac{3}{6}=\dfrac{1}{2}[/tex]
Then, the probability that at least one of the two systems tested will be defective will be :_
[tex]P(x\geq1)=1-P(x=0)\\\\=1-^{2}C_{0}(\dfrac{1}{2})^0(\dfrac{1}{2})^{2} \ \ \text{[By using formula (1)]}\\\\=1-(\dfrac{1}{2})^2\ \ \because\ ^nC_0=1\\\\=1-\dfrac{1}{4}=\dfrac{3}{4}[/tex]
Hence, the probability that at least one of the two systems tested will be defective = [tex]\dfrac{3}{4}[/tex]
b) If four of the six systems are actually defective.
[tex]\Rightarrow\ p=\dfrac{4}{6}=\dfrac{2}{3}[/tex]
Then, the probability that at least one of the two systems tested will be defective will be :_
[tex]P(x\geq1)=1-P(x=0)\\\\=1-^{2}C_{0}(\dfrac{2}{3})^0(1-\dfrac{2}{3})^{2} \ \ \text{[By using formula (1)]}\\\\=1-(\dfrac{1}{3})^2\ \ \because\ ^nC_0=1\\\\=1-\dfrac{1}{9}=\dfrac{8}{9}[/tex]
Hence, the probability that at least one of the two systems tested will be defective =[tex]\dfrac{8}{9}[/tex]
The probability that both are defective :
[tex]P(x=2)=^{2}C_{2}(\dfrac{2}{3})^2(1-\dfrac{2}{3})^{0} \ \ \text{[By using formula (1)]}\\\\=(\dfrac{2}{3})^2\ \ \because\ ^nC_n=1\\\\=\dfrac{4}{9}[/tex]
Hence, the probability that both are defective = [tex]\dfrac{4}{9}[/tex]
