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Poisson call arrival probabilities

Short answer

Arrival rate λ=0.2 calls/s, interval t=10 s, mean λt=2. Poisson probabilities: no call P(0)=e^{-2}=0.1353; one call P(1)=0.2707; two calls P(2)=0.2707; more than two calls P(>2)=1−(0.1353+0.2707+0.2707)=0.3233.

On average, one call arrives every 5 seconds. During a period of 10 seconds, what is the probability that: i) No call arrives? ii) One call arrives? iii) Two calls arrive? iv) More than two calls arrive?

Random call arrivals follow the Poisson distribution: the probability of exactly k calls in time t is

P(k)=(λt)kk! e−λtP(k) = \frac{(\lambda t)^k}{k!}\,e^{-\lambda t}

Given: one call every 5 s on average, so λ=0.2\lambda = 0.2 calls/s; t = 10 s.

λt=0.2×10=2\lambda t = 0.2 \times 10 = 2

(i) No call

P(0)=e−2=0.1353P(0) = e^{-2} = 0.1353

(ii) One call

P(1)=211!e−2=2×0.1353=0.2707P(1) = \frac{2^1}{1!}e^{-2} = 2 \times 0.1353 = 0.2707

(iii) Two calls

P(2)=222!e−2=2×0.1353=0.2707P(2) = \frac{2^2}{2!}e^{-2} = 2 \times 0.1353 = 0.2707

(iv) More than two calls

P(k>2)=1−[P(0)+P(1)+P(2)]=1−(0.1353+0.2707+0.2707)=1−0.6767=0.3233\begin{aligned} P(k>2) &= 1 - [P(0) + P(1) + P(2)] \\ &= 1 - (0.1353 + 0.2707 + 0.2707) \\ &= 1 - 0.6767 = 0.3233 \end{aligned}
kP(k)
00.1353
10.2707
20.2707
> 20.3233

Answer: (i) 0.1353, (ii) 0.2707, (iii) 0.2707, (iv) 0.3233.

✓ Answer checked by Claude Opus 5.5 (high effort)

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