Question

Question 8 Try again. Hint: "More than four (4) " means" (1- Probability (up to and including four (4)). Traffic engineers install 6 street lights with new bulbs. The probability that a bulb will fail within 50040 hours of operation is 0.15. Assume that each of the bulbs fails independently (a) What is the probability that fewer than two of the original bulbs will fail within 50040 hours of operation? (b) What is the probability that no bulbs will have to be replaced within 50040 hours of operation? (c) What is the probability that more than four of the original bulbs will need replacing within 50040 hours? (a .w..w.. Your answer should be correct to 3 decimal places. . Your answer should be correct to 3 decimal places. Yo answer should be correct to 6 decimal places.

Answer 1

a) here pth percentile =-ln(1-p)/$\lambda$=-ln(1-0.15)/$\lambda$ =50040

$\lambda$ =0.00000325 faiilure/hour

a)

expected number of failure in 50040 hours=50040*0.00000325 =0.162519

hence P(fewer than 2 bulb fail) =P(X<=1)=P(X=0)+P(X=1)=0.988

b)P(no bulb fail)=P(X=0)=e^-0.162519*0.162519⁰/0! =0.85

c)P(X>4)=1-P(X<=4)=1-0.999999 =0.000001