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The time until the next car accident for a particular driver is exponentially distributed with a mean of 200 days. Calculate the probability that the driver has no accidents in the next 365 days, but then has at least one accident in the 365-day period that follows this initial 365-day period.

Attempt

Let $T$ be the time it takes for a driver to have a car accident. We are given $T$ is $exp( \lambda = 1/200 )$. We need to find

$$ P(T > 365) = 1 - F(365) = 1 - 1 + e^{-365/200} = 0.1612 $$

Is this correct? MY answer key says the correct answer should be $\boxed{0.1352}$. What am I missing here?

Community
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James
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3 Answers3

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You want the first accident to be between the first year and second year.

\begin{align} P(365< T \leq 2 \cdot 365) &= F(2 \cdot 365) - F(365) \end{align}

Siong Thye Goh
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Indeed if you read the problem carefully you want: $$P(365 * 2 >T > 365 ) = F(365*2) - F(365)$$

Frostic
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  • This is not enough for an answer. Reading carefully is not doing maths. Maths is reasoning. So if you want to answer this question in a useful way, please explain every reasoning from the problem statement to the result. – Tom-Tom Apr 23 '18 at 19:46
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You want to go the next 365 days accident-free and then want to have an accident within the next 365 days (days 365-730). After day 730 you do not want any more accidents.

Day 0-365: No accidents

Day 365-730: At least 1 accident

Day 730+: No accidents

In the solution, you forgot to subtract off the probability of an accident occurring after day 730. After doing so you get

$$P(730>T>365)=F(730)−F(365)=e^{730/200}-e^{365/200}=0.13523$$

Maxwell Johnson
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