Answer :
Answer:
1. 0.0000454
2. 0.01034
3. 0.0821
4. 0.918
Step-by-step explanation:
Let X be the random variable denoting the number of passengers arriving in a minute. Since the mean arrival rate is given to be 10,
[tex] X \sim Poi(\lambda = 10) [/tex]
1. Requires us to compute
[tex] P(X = 0) = e^{-10} \frac{10^0}{0!} = 0.0000454[/tex]
2. We need to compute [tex] P(X \leq 3) = P(X =0) + P(X =1) + P(X =2) + P(X =3) [/tex]
[tex] P(X =1) = e^{-10} \frac{10^1}{1!} = 0.000454 [/tex]
[tex] P(X =2) = e^{-10} \frac{10^2}{2!} = 0.00227 [/tex]
[tex] P(X =3) = e^{-10} \frac{10^3}{3!} = 0.00757 [/tex]
[tex]P(X \leq 3) =0.0000454+ 0.000454 + 0.00227 + 0.00757 = 0.01034[/tex]
3. The expected no. of arrivals in a 15 second period is = [tex] 10 \times \frac{1}{4} = 2.5 [/tex]. So if Y be the random variable denoting number of passengers arriving in 15 seconds,
[tex] Y \sim Poi(2.5) [/tex]
[tex] P(Y=0) = e^{-2.5} \frac{2.5^0}{0!} = 0.0821[/tex]
4. Here we use the fact that Y can take values [tex] 0,1, \dotsc [/tex]. So, the event that "Y is either 0 or [tex]\geq[/tex] 1" is a sure event ( i.e it has probability 1 ).
[tex]P(Y=0) + P(Y \geq 1) = 1 \implies P(Y \geq 1) = 1 -P(Y=0) = 1 - 0.0821 = 0.918[/tex]