Answer to Question #301253 in Statistics and Probability for Uncle B

Question #301253

In a box of switches, it is known that 10% of the switches are faulty. A technician is wiring 30 circuits,



each of which needs one switch. What is the probability that (a) all 30 work (b) at most two of the



switches do not work.



2

1
Expert's answer
2022-02-23T12:26:18-0500

Let the random variable XX represent the number of faulty switches then, XBinomial(n=30,p=0.1)X\sim Binomial (n=30,p=0.1) given as,

p(X=x)=(30x)0.1x0.930x, x=0,1,2,3,.....,30p(X=x)=\binom{30}{x}0.1^x0.9^{30-x}, \space x=0,1,2,3,.....,30


a)

To find the probability that all 30 switches work is equivalent to finding the probability that there are no faulty switches. This can be written as,

p(X=0)=(300)0.100.930=0.930=0.04239116p(X=0)=\binom{30}{0}0.1^00.9^{30}=0.9^{30}=0.04239116

The probability that all 30 switches work is 0.04239116


b)

p(X2)=p(x=0)+p(x=1)+p(x=2)=((300)0.100.930)+((301)0.110.929)+((302)0.120.928)=0.04239116+0.1413039+0.2276562=0.4113512p(X\le 2)=p(x=0)+p(x=1)+p(x=2)=(\binom{30}{0}0.1^00.9^{30})+(\binom{30}{1}0.1^10.9^{29})+(\binom{30}{2}0.1^20.9^{28})=0.04239116+ 0.1413039+ 0.2276562= 0.4113512

The probability that there are at most 2 faulty switches is 0.4113512


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