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Poisson arrival probabilities in 15 minutes

Short answer

With mean m = 3 vehicles in 15 minutes, exactly 3 has probability 0.2240, less than 3 has probability 0.4232, and more than 3 has probability 0.3528.

The vehicle arrivals at the section of road are assumed to be Poisson distributed with an average arrival rate of 1 vehicle every 5 minutes. What is the probability of (i) Exactly 3 vehicles arrive in a 15 minute interval (ii) less than 3 vehicles arrive in a 15 minute interval? (iii) More than 3 vehicles arrive in a 15 minute intervals?

For Poisson arrivals, the probability of nn arrivals in an interval is P(n)=mne−mn!P(n) = \dfrac{m^n e^{-m}}{n!} where m=λtm = \lambda t is the mean number of arrivals in the interval.

Rate: 1 vehicle per 5 min, so in 15 min m=15/5=3m = 15/5 = 3 vehicles.

(i) Exactly 3 vehicles

P(3)=33e−33!=27(0.04979)6=0.2240P(3) = \frac{3^3 e^{-3}}{3!} = \frac{27(0.04979)}{6} = 0.2240

(ii) Less than 3 vehicles

P(n<3)=P(0)+P(1)+P(2)=0.0498+0.1494+0.2240=0.4232\begin{aligned} P(n<3) &= P(0)+P(1)+P(2)\\ &= 0.0498 + 0.1494 + 0.2240 = 0.4232 \end{aligned}

(iii) More than 3 vehicles

P(n>3)=1−[P(0)+P(1)+P(2)+P(3)]=1−(0.4232+0.2240)=0.3528P(n>3) = 1 - [P(0)+P(1)+P(2)+P(3)] = 1 - (0.4232 + 0.2240) = 0.3528

Answer: (i) 0.2240 (22.4 %), (ii) 0.4232 (42.3 %), (iii) 0.3528 (35.3 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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