Question
A bus station has exactly four entrances. In the morning the numbers of passengers arriving at these entrances during a 10-second period have the independent distributions $Po(0.4)$, $Po(0.1)$, $Po(0.2)$ and $Po(0.5)$.
Find the probability that the total number of passengers arriving at the four entrances to the bus station during a randomly chosen 1-minute period in the morning is more than 3.
▶️Answer/Explanation
solution:-
Step 1: Define the Poisson Distributions
Each entrance follows an independent Poisson distribution for a 10-second period:
The total number of arrivals in 10 seconds is:
Since the sum of independent Poisson distributions is also Poisson, we get:
Step 2: Scale to 1 Minute (60 Seconds)
Since Poisson processes scale linearly with time, in 60 seconds (which is 6 times a 10-second period), the total number of passengers follows:
We need to find:
Step 3: Compute
The Poisson cumulative probability formula is:
where
.
We calculate:
Approximating using a calculator:
Summing these:
Thus:
Final Answer:
\(0.928\) (rounded to 3 decimal places)
Markscheme———–
$\lambda=7.2$
$P(X>3)=1-e^{-7.2}(1+7.2+\frac{7.2^{2}}{2!}+\frac{7.2^{3}}{3!})$
or $1-e^{-7.2}(1+7.2+25.92+62.21)$
or $1-(0.0007466+0.005375+0.01935+0.04644)$
= 0.928 (3sf)