NASA and rare planet transits. A “planet transit” is a rare celestialevent in which a planet appears to cross in front ofits star as seen from Earth. The National Aeronautics andSpace Administration (NASA) launched its Kepler mission, designed to discover new planets in the MilkyWay by detecting extrasolar planet transits. After 1 year of themission in which 3,000 stars were monitored, NASA announcedthat five-planet transits had been detected(NASA, American Astronomical Society, January 4, 2010).Assume that the number of planet transits discovered for every3,000 stars follows a Poisson distribution with. What is the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 10 planettransits willbe seen?

The number of planet transit discovered for every 3000 stars follows a Poisson distribution with$\lambda =5$

The probability that X represents the number of successes of a random variable in Poisson process is given by:

$P\left(X=x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

Where,

X is the number of successes

$e=2.71828$is the Euler number

$\lambda$is the mean in the given time interval.

We have $\lambda =5$

We have to calculate the probability that, in the next 3,000 stars monitored by the Kepler mission, more than 10 planet transits will be seen that is $P\left(X>10\right)$.

We can see either 10 or less than 10 planets, or we can see more than 10 planets.

We know that the sum of all the probabilities is 1.

Hence,

$$\begin{gathered} P\left(X\le 10\right)+P\left(X>10\right)=1 \\ P\left(X>10\right)=1-P\left(X\le 10\right) \end{gathered}$$

Where,

$$\begin{gathered} P\left(X\le 10\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)+ \\ P\left(x=5\right)+P\left(x=6\right)+P\left(x=7\right)+P\left(x=8\right)+P\left(x=9\right)+P\left(x=10\right) \end{gathered}$$

Hence, from the formula of Poisson distribution:

$$\begin{gathered} P\left(X=x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!} \\ P\left(X=0\right)=\frac{e^{-5}{\lambda }^0}{0!} \\ =0.0067 \\ P\left(X=1\right)=\frac{e^{-5}{\lambda }^1}{1!} \\ =0.0337 \\ P\left(X=2\right)=\frac{e^{-5}{\lambda }^2}{2!} \\ =0.0842 \\ P\left(X=3\right)=\frac{e^{-5}{\lambda }^3}{3!} \\ =0.1404 \\ P\left(X=4\right)=\frac{e^{-5}{\lambda }^4}{4!} \\ =0.1755 \\ P\left(X=5\right)=\frac{e^{-5}{\lambda }^5}{5!} \\ =0.7154 \\ P\left(X=6\right)=\frac{e^{-5}\lambda 6}{6!} \\ =0.1462 \\ P\left(X=7\right)=\frac{e^{-5}{\lambda }^7}{7!} \\ =0.1044 \\ P\left(X=8\right)=\frac{e^{-5}{\lambda }^0}{0!} \\ =0.0067 \\ P\left(X=9\right)=\frac{e^{-5}{\lambda }^9}{9!} \\ =0.0362 \\ P\left(X=10\right)=\frac{e^{-5}{\lambda }^{10}}{10!} \\ =0.0181 \end{gathered}$$

Adding all the probabilities,

$$\begin{gathered} P\left(x\le 10\right)=0.0067+0.0337+0.0842+0.1404+0.1755+0.1754+ \\ 0.1460+0.1043+0.06520+0,0362+0.0181 \\ P\left(x\le 10\right)=1-P\left(x\le 10\right) \\ =1-0.9658 \\ =0.0342 \end{gathered}$$

Hence, there is 3.4% probability that, in the next 3000 stars monitored by the Kepler mission, more than 4 planet transits will be seen .