1. ## Distribution of Poison

A dealership sells an average of 0.5 cars per day. The company needs to pay a debt that matures in two days. Must sell at least three cars in two days, what is the likelihood that the company can pay the debt?

2. If the company sells 0.5 cars in one day, how many cars do they sell over two days?

Once you make that transformation, it is simply a matter of plugging and chugging into the formula for the Poisson distribution.

3. Originally Posted by ANDS!
If the company sells 0.5 cars in one day, how many cars do they sell over two days?

Once you make that transformation, it is simply a matter of plugging and chugging into the formula for the Poisson distribution.
Ok. I find:

$\displaystyle \frac{e^{-1} 1^3}{3!} = 0,06131$
But the answer is: $\displaystyle 0,0803$

4. Well the company must sell at LEAST 3 cars. Not exactly 3 cars. The probability that they sell exactly three cars is 0.0613. How would you go about finding the probability that they sell at LEAST 3 cars; we are still using Poisson here, but you're going to have to take the compliment of something.

5. Originally Posted by ANDS!
Well the company must sell at LEAST 3 cars. Not exactly 3 cars. The probability that they sell exactly three cars is 0.0613. How would you go about finding the probability that they sell at LEAST 3 cars; we are still using Poisson here, but you're going to have to take the compliment of something.
I understand. Can sell more than three cars. But I do not know how to calculate this probability

6. Well if they can sell more than 3 cars, then theoretically they can sell 100 cars or 1000 cars or "n" number of cars. So to find out the Poisson Distribution of "at least 3 cars. . ." you would have to calculate Poisson(3)+Poisson(4)+Poisson(5)+. . .Poisson(n). Of course this isn't possible (or how you're supposed to do this). Thus you need another alternative.

As I said, you can use the idea of the compliment of an event. If they don't sell at LEAST 3 cars in two days, then how many cars at MOST did they sell. Find the Poisson Distribution of the AT MOST statement, and see if something clicks.

7. Originally Posted by ANDS!
Well if they can sell more than 3 cars, then theoretically they can sell 100 cars or 1000 cars or "n" number of cars. So to find out the Poisson Distribution of "at least 3 cars. . ." you would have to calculate Poisson(3)+Poisson(4)+Poisson(5)+. . .Poisson(n). Of course this isn't possible (or how you're supposed to do this). Thus you need another alternative.

As I said, you can use the idea of the compliment of an event. If they don't sell at LEAST 3 cars in two days, then how many cars at MOST did they sell. Find the Poisson Distribution of the AT MOST statement, and see if something clicks.
I was adding: P(3) + P(4) + P(5) + P(6) + .....
I checked that was approaching to 0,0803. But, how can I be sure? I find answer $\displaystyle < 0,0803$ and answer $\displaystyle > 0,0803$

8. Do this for me, find that Probability that they sell at most 2 cars (meaning they might sell 0, 1 or 2 cars). Let me know what you get.

9. Originally Posted by ANDS!
Do this for me, find that Probability that they sell at most 2 cars (meaning they might sell 0, 1 or 2 cars). Let me know what you get.
Knowing the probability of paying the debt. To pay the debt has to sell at least 2 cars, then P (> = 2)

How to get an exact answer, if P(2) + P(3) + .... ???

10. Apprentice, you know the Poisson formula. Add them up from 0 to 2 and subtract from 1.

They say at least 3 must be sold. That can be 3 up to however many.

But, since there is no given bound, find the probabilties from 0 to 2 and subtract from 1.

$\displaystyle 1-\sum_{k=0}^{2}\frac{e^{-1}}{k!}$

Because $\displaystyle {\lambda}=.5(2)=1$

11. Originally Posted by galactus
Apprentice, you know the Poisson formula. Add them up from 0 to 2 and subtract from 1.

They say at least 3 must be sold. That can be 3 up to however many.

But, since there is no given bound, find the probabilties from 0 to 2 and subtract from 1.

$\displaystyle 1-\sum_{k=0}^{2}\frac{e^{-1}}{k!}$

Because $\displaystyle {\lambda}=.5(2)=1$
Ohhh yes. Thank you

12. depends on how much poison you take