The probability that she would get at least 20,000 additional miles out of it is 0.3678.
Let X denote the total number of thousands of miles that the car is driven before it needs to be junked. We want to compute P({X≥30} | {X≥10}).
Assuming X is an exponential random variable with parameter 1/20, we have
P{X≥30} = [tex]e^{-3/2}[/tex]
P{X≥10} = [tex]e^{-1/2}[/tex]
P({X≥30} | {X≥10}) = P{X≥30} / P{X≥10}
= [tex]e^{-3/2}[/tex] / [tex]e^{-1/2}[/tex]
= [tex]e^{-1}[/tex]
≈ 0.3678
Hence, the probability is 0.3678.
To learn more about probability here:
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