Clearification on Poisson Distribution

**lllll** · March 23rd 2008, 10:18 PM

Q: A parking lot has 2 entrances. Cars arrive at entrance 1 and 2 according to a Poisson distribution an average of 3/hour, and entrance 2 at an average of 4/hour, respectively. What is the probability that a total of 3 cars will arrive at the parking lot in a given hour? (assume arrivals are independent at the 2 entrances).

Solution:

$\begin{array}{l} \lambda_p = \lambda_1 + \lambda_2 = 3+4 = 7 \\ \\ P(3) = \frac{7^3}{3!} e^{-7} =0.0521295\\ \end{array}$

Is this correct?

**CaptainBlack** · March 23rd 2008, 10:52 PM

Originally Posted by lllll

Q: A parking lot has 2 entrances. Cars arrive at entrance 1 and 2 according to a Poisson distribution an average of 3/hour, and entrance 2 at an average of 4/hour, respectively. What is the probability that a total of 3 cars will arrive at the parking lot in a given hour? (assume arrivals are independent at the 2 entrances).

Solution:

$\begin{array}{l} \lambda_p = \lambda_1 + \lambda_2 = 3+4 = 7 \\ \\ P(3) = \frac{7^3}{3!} e^{-7} =0.0521295\\ \end{array}$

Is this correct?

Yes

RonL

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