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Here is the problem.

A car factory has an accident rate of 0.3 per month.

a) Find after how many months we expect at least one accident with a probability more than 85%. b) To motivate its workers, the company pays the union a bonus of 5000USD if no accidents occur in a given month and 1000USD if there is 1 or 2 per month. Find b) the expected money the company will have to pay for a given month.

For a) I got it, basically $P(X\ge 1)=0.85->1-P(X=0)=0.85->1-e^\lambda=0.85->\lambda=0.82$ That value corresponds somewhere between 2 and 3 months, so after 2 months of operation we expect an accident with a 85% probability during the 3rd month.

Could you give me any hint for b) I thought about doing something Money to be paid as bonus $5000*P(X=0)+ 1000(P(X=1)+P(X=2))$ but I am not sure it goes in the right direction.

Just_Newbie
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1 Answers1

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Define $X$ to be the payment by the company in one month. The number of accidents in one month is distributed $\text{Poisson}(.3)$. Define a variable $Y$ that is the number of accidents per month, then the event that $X=5000$ is the same as $Y=0$ and $X=10000$ is the same as $Y=0\text { or } 1$. So using the definition of expected value

$$\begin{split}E(X)&=5000 P(X=5000)+10000P(X=10000)\\ &=5000P(Y=0)+10000\left(P(Y=1)+P(Y=2)\right)\\ &=5000(.7408)+10000(.2222+.0333)\\ &=6259\end{split}$$

Note: In part a the poisson process can be regarded continually, so the $\lambda=.82$ corresponds to $.82/.3=2.733333$ months, so after about $2.73333$ months the probability of at least one accident will be at least .85. This is about one month and 23 days.

Vons
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  • what I don't get, why do you put the bonus amount in the $P$ i.e. like $5000P(X=5000)$ ? – Just_Newbie May 10 '21 at 16:59
  • That is the definition of expected value for a discrete random variable. $E(X)=\sum_i x_i P(X=x_i)$. But I have added an edit to make it clearer, you have the right process. – Vons May 10 '21 at 17:41