Answer:
There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.
Step-by-step explanation:
Assume that the number of planet transits discovered for every 3,000 stars follows a Poisson distribution with λ=5.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Clambda%7D%2A%5Clambda%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
is the Euler number
is the mean in the given time interval.
For this problem, we have that ![\lambda = 5](https://tex.z-dn.net/?f=%5Clambda%20%3D%205)
What is the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen?
That is
. We either see 4 or less planets, or we see more than 4. The sum of the probabilities is decimal 1. So
![P(X \leq 4) + P(X > 4) = 1](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%2B%20P%28X%20%3E%204%29%20%3D%201)
![P(X > 4) = 1 - P(X \leq 4)](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29)
In which
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
So
![P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Clambda%7D%2A%5Clambda%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-5}*5^{0}}{(0)!} = 0.0067](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-5%7D%2A5%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.0067)
![P(X = 1) = \frac{e^{-5}*5^{1}}{(1)!} = 0.0337](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-5%7D%2A5%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.0337)
![P(X = 2) = \frac{e^{-5}*5^{2}}{(2)!} = 0.0842](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-5%7D%2A5%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.0842)
![P(X = 3) = \frac{e^{-5}*5^{3}}{(3)!} = 0.1404](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-5%7D%2A5%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.1404)
![P(X = 4) = \frac{e^{-5}*5^{4}}{(4)!} = 0.1755](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-5%7D%2A5%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.1755)
![P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 = 0.4405](https://tex.z-dn.net/?f=P%28X%20%5Cleq%204%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.0067%20%2B%200.0337%20%2B%200.0842%20%2B%200.1404%20%2B%200.1755%20%3D%200.4405)
Finally
![P(X > 4) = 1 - P(X \leq 4) = 1 - 0.4405 = 0.5595](https://tex.z-dn.net/?f=P%28X%20%3E%204%29%20%3D%201%20-%20P%28X%20%5Cleq%204%29%20%3D%201%20-%200.4405%20%3D%200.5595)
There is a 55.95% probability that, in the next 3,000 stars monitored by the Kepler mission, more than 4 planet transits will be seen.