How to find the probability there are 11 or more cars? During rush hour the number of cars passing through a particular intersection has a Poisson distribution with an average of 540 per hour.
Find the probability there are 11 or more cars?
The answer is 0.006669.
I don not know how to deal with this kind of question.
Please help.Thank you very much.
 A: Let $X$ denote the random variable of the number of cars passing through the intersection in the time interval of $T$ hours. The probability of exactly $n$ cars crossing the intersection equals:
$$
   \Pr(X=n) = \frac{(\mu T)^n}{n!} \mathrm{e}^{-\mu T}
$$
where $\mu = 540 \frac{\text{cars}}{\text{hour}}$. The event that $A = \{X \geqslant 11\}$ is complementary to $B = \{0 \leqslant X \leqslant 10\}$, and hence $\Pr(A) + \Pr(B) = 1$:
$$
    \Pr(X \geqslant 11) = 1-\Pr(0 \leqslant X \leqslant 10) = 1 - \sum_{m=0}^{10} \Pr(X=m)
$$
A: I think you mean that the probability is 1 - 0.006669?
I'm assuming you're talking about the probability of 11 or more cars in an hour.
The probability of 1 or more cars is simply:
$$1 - exp(-540).$$
The probability of 2 or more cars is:
$$1 - exp(-540) \cdot (1 + 540).$$
The probability of 11 or more cars is:
$$1 - exp(-540) \cdot \sum_{i=0}^{10}\frac{540^{i}}{i!}.$$
This answer is going to be very close to 1, which intuitively makes sense.
A: Let $X$ be a Poisson random variable that describes the number of cars in one hour with $\lambda = 540$. Then:
$P(X\ge11)=1-P(X<11)=1-P(X=10)-P(X=9)-...-P(X=0)$
where $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$
