Question 1175123: An important part of the customer service responsibilities of a cable company relates to the speed with which
trouble in service can be repaired. Historically, the data show that the likelihood is 0.85 that troubles in a
residential service can be repaired on the same day. Use the binomial formula to calculate.
a. For the first five troubles reported on a given day, what is the probability that: All five will be repaired on
the same day?
b. For the first five troubles reported on a given day, what is the probability that: Fewer than two troubles
will be repaired on the same day?
c. For the first five troubles reported on a given day, what is the probability that: At least 3 troubles will be
repaired on the same day.
d. Find the mean number of troubles repaired on the same day?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is 0.85^5=0.4437
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fewer than 2 is 1 or 0
probability of 0 fixed is 0.15^5=0.00008 plus p(1)=5C1*0.85^1*0.15^4=0.0022. The sum is 0.0230 prob.
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at least 3: probability of 2 being fixed is 5C2*0.85^2*0.15^3=0.0244. Add that to the above answer and get 0.0266. At least 3 is 1- prob of (0, 1, 2) or 0.9734. Can use calculator to get right sided probability by doing 1-probability of the left side, so 1-binomcdf(5,.85,2) will give the probability of greater than 2, which is the same as at least 3.
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mean is np=4.25 troubles.
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