Question

Suppose that a long distance taxi service owns 4 vehicles. These are of different ages and have different repair records. The probabilities that, on a given day, each vehicle will be available for use are: 0.90, 0.90, 0.80, 0.70. Whether one vehicle is available is independent of whether any other vehicle is available. a. Find the probability distribution for the number of vehicles available for use on a given day. b. Find the expected number of vehicles available for use on a given day. c. Find the standard deviation of the number of vehicles available for use on a given day. Round your answer to four decimal points

Answer 1

a) Let X be the number of vehicles that will be available for use.

Probability tha all 4 will be available for use is

P(X=4) = 0.90*0.90*0.80*0.70 = 0.4536

Probability that 3 cars will be available for use ( 3 car can be any of the 4 cars) is

P(X = 3) = 0.90*0.90*0.80*(1-0.70) + 0.90*0.90*(1-0.80)*0.70 + 0.90*(1-0.90)*0.80*0.70 + (1-0.90)*0.90*0.80*0.70 = 0.4086

Probability that 2 cars will be available for use ( 2 car can be any of the 4 cars) is

P(X=2) = 0.9*0.9*(1-0.80)*(1-0.70) + 0.9*(1-0.9)*0.8*(1-0.7) + (1-0.9)*0.9*0.8*(1-0.7) + 0.9*(1-0.9)*(1-0.8)*0.7 + (1-0.9)*0.9*(1-0.8)*0.7 + (1-0.9)*(1-0.9)*0.8*0.7

P(X=2) = 0.1154

Probability that 1 cars will be available for use ( 1 car can be any of the 4 cars) is

P(X=1) = 0.9*(1-0.9)*(1-0.80)*(1-0.70) + (1-0.9)*0.9*(1-0.8)*(1-0.7) + (1-0.9)*(1-0.9)*0.8*(1-0.7) + (1-0.9)*(1-0.9)*(1-0.8)*0.7

P(X=1) = 0.0146

Probability that 0 cars will be available for use is

P(X = 0) = (1-0.9)*(1-0.9)*(1-0.80)*(1-0.7) = 0.0006

X	0	1	2	3	4
P(X)	0.0006	0.0146	0.1154	0.4086	0.4536

b)

The expected number of vehicles available for use on a given day =

c)

the standard deviation of the number of vehicles available for use on a given day =

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