For a Poisson distribution model, if arrival rate of passengers at an airport is recorded 30 per hour on a given day. Find the probability of 4 or fewer arrivals in the first 10 minutes of an hour.

0.438

The correct answer is Option (1) → 0.438

Given number of arrivals per hour is 30.

Let the random variable X be the number of arrivals in first 10 minutes of an hour i.e. $\frac{1}{6}$ hour. 

$∴ E(X) = 30 ×\frac{1}{6} =5$

the expected number of arrivals in the first 10 minutes of an hour is 5.

P(4 or fewer arrivals in first 10 minutes of an hour)

$= P(X ≤4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$=\frac{5^0e^{-5}}{0}+\frac{5^1e^{-5}}{1}+\frac{5^2e^{-5}}{2}+\frac{5^3e^{-5}}{3}+\frac{5^4e^{-5}}{4}$

$=e^{-5}\left[1+5+\frac{25}{2}+\frac{125}{6}+\frac{625}{24}\right]=0.0067×\frac{1569}{24}=0.438$