3
$\begingroup$

I am trying to find the probability for a Poisson distribution. The mean is two cars sold per day. The question is:

"What is the probability that at least one car is sold for each of three consecutive days?"

I know that the probability of at least one car being sold for $1$ day is:

$P$($X$ $≤$ $1$) $ =$ $P$($X = 1$) + $P$($X = 0$) = $0.135335283 + 0.270670566$ = $0.406005849$

But the part that throws me off is the term "for EACH of three consecutive days". If the question was "find the probability of at least one car being sold for three days", all I would have to do it multiply the mean ($2$ cars) by $3$ days and the final answer would be $0.017351265$.

But since when the question says "for EACH of three consecutive days", does it mean I take the probability of at least one car sold in 1 day, and multiply it by itself for each of the three days? That is: $0.406005849$ to the power of $3$ = $.066926308$.

I just want to know what is the correct way to calculate it by "each consecutive day." Should the answer be $0.017351265$ or $.066926308$. Any help would be appreciated.

  • 0
    Your expression for the probability of at least one car being sold for one day is not right. It shouldn't be $P(X \leq 1)$, as that is the probability of at *most* one car being sold. You want $P(X \geq 1)$, which can be written as $1-P(X = 0)$. ¶ You might ask yourself if your value makes sense. If I average two cars per day, does it make sense that I have a less than $1/2$ probability of selling at least one car? And so on.2016-03-08
  • 0
    And then Gerry Myerson's approach is fastest: Just cube $P(X \geq 1)$ and you will obtain your answer. It doesn't matter that the three days are consecutive, incidentally.2016-03-08

2 Answers 2