In: Statistics and Probability
A lap top computer manufacturer tests each device before leaving the factory. From history we know that the probability of failure is 0.9. If four laptops are randomly selected:
(1) what is the probability of fewer than 2 laptops fail the inspection?
a. 0.0037
b. 0.0058
c. 0.0074
d. 0.0024
(2) what is the probability that the third laptop which fails is the 4th selected for inspection?
a. 0.2187
b. 0.7516
c. 0.3758
d. 0.4374
1.Let X be the random variable representing the number of failed laptops. Then X can follow binomial distribution with n=4 and p=0.9.then q=1-p=0.1
.probability that fewer than two laptops fail the inspection =P(X<2)=P(X=0,1)=P(X=0)+P(X=1){ As the events are independent} =4C0×(p^0)×(q^4) + 4C1×p×q^3, where C represents combinatorial function. = 1×1×(0.1^4)+ 4×0.9×0.001=0.0001+0.0036=0.0037.So the right option is option A.
2.3rd laptop fails in fourth selection that means 2 laptop fail in first three selections. So we are sure that 4th selection was failed selection or the laptop at 4th slection. If we define X as random variable defining number of laptops that failed inspection ,since here ordering is also important so X will follow negative binomial distribution with x=1, r=3, p=0.9 ,q=0.1
Then P(X=1)=(x+r-1)Cx × p^r ×q^x-1 =(1+3-1) C1×0.9^3×0.1=3×0.9^3 ×0.1 =0.2187. So the correct answer is option A.
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