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David is deciding to buy either a new car or a used car that has been driven 10,000 miles. For each car, find the probability that he can drive it for an extra 20,000 miles in the following scenarios.

(a) The total mileage that both cars can be driven before they completely break down is an exponential random variable with mean 20 thousands miles.
(b) The total mileage that both cars can be driven before they completely break down is uniformly distributed over (0, 40,000).

Respuesta :

Answer:

a) [tex]P(X>30000 | X>10000)=0.3679[/tex]

b) [tex]P(X>30000 | X>10000)=0.5[/tex]

Step-by-step explanation:

a) In the given case,

Mileage of the cars = [tex]X \sim E X p(20000)[/tex]

The PDF is [tex]f(x)=\frac{1}{20000} e^{-x / 20000}, x>0[/tex]

The CDF is [tex]F(x)=1-e^{-x / 20000}, x>0[/tex]

For new car the probability will be

[tex]P(X>20000)=1-F(20000)[/tex]

[tex]P(X>20000)=e^{-20000 / 20000}[/tex]

[tex]P(X>20000)=0.3679[/tex]

For the used car probability is

[tex]P(X>30000 | X>10000)=\frac{P(X>30000 \cap X>10000)}{P(X>10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{P(X>30000)}{P(X>10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{1-F(30000)}{1-F(10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{e^{-30000 / 20000}}{e^{-10000 / 20000}}[/tex]

[tex]P(X>30000 | X>10000)=0.3679[/tex]

b) For the case given in this part the mileage of the cars is [tex]X \sim[/tex] Uniform (0,40000)

The PDF is [tex]f(x)=\frac{1}{40000}, x>0[/tex]

The CDF is [tex]F(x)=\frac{x}{40000}, x>0[/tex]

For the new car probability is [tex]P(X>20000)=1-F(20000)[/tex]

[tex]P(X>20000)=1-\frac{20000}{40000}[/tex]

[tex]P(X>20000)=0.5[/tex]

For the old car probability is, [tex]P(X>30000 | X>10000)=\frac{P(X>30000 \cap X>10000)}{P(X>10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{P(X>30000)}{P(X>10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{1-F(30000)}{1-F(10000)}[/tex]

[tex]P(X>30000 | X>10000)=\frac{1-\frac{30000}{40000}}{1-\frac{20000}{40000}}[/tex]

[tex]P(X>30000 | X>10000)=0.5[/tex]

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