# Poisson Distribution for Consecutive Figures

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.

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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. – Brian Tung Mar 8 at 19:13
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. – Brian Tung Mar 8 at 19:14

You have to make the assumption that the number of cars sold on day $i$ is independent of the number of cars sold on day $j$, where $i\neq j$.

The first step is to figure out the probability of at least one car being sold on a single day. You're math above is correct, but you've incorrectly interpreted "at least one". If $X_i$ is the number of cars sold on day $i$, then you want to find $P(X_i\geq1)=1-P(X_i=0)$. With the given Poisson distribution, $P(X_i\geq1)=1-P(X_i=0)=1-e^{-2}$. Since we've done this for a general day, it is the probability of selling at least one car for any day.

Now, you want to find $P(X_1\geq1,X_2\geq1,X_3\geq1)$. We can rewrite this probability using conditioning: $$P(X_1\geq1,X_2\geq1,X_3\geq1)=P(X_1\geq1)P(X_2\geq1\mid X_1\geq1)P(X_3\geq1\mid X_1\geq1,X_2\geq1)$$ By assuming that the number of cars sold on each day is independent of the number of cars sold on the other days, the probability simplifies to $$P(X_1\geq1,X_2\geq1,X_3\geq1)=P(X_1\geq1)P(X_2\geq1)P(X_3\geq1)=(1-e^{-2})^3$$

Another way to think about the problem is this:

Each day you flip a weighted coin. If the coin lands tails, you sell no cars. If the coin lands heads, you sell at least one car. The probability of the coin landing heads is $1-e^{-2}$ and the probability of the coin landing tails is $e^{-2}$. For three consecutive days you flip this coin. What is the probability of three heads?

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If the probability of at least one car being sold is $p$, then the probability of this happening three days in a row is $p\times p\times p$, which is $p^3$.

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