Answer:
The probability that at least 1 car arrives during the call is 0.9306
Explanation:
Cars arriving according to Poisson process - 80 Cars per hour
If the attendant makes a 2 minute phone call, then effective λ = 80/60 * 2 = 2.66666667 = 2.67 X ≅ Poisson (λ = 2.67)
Now, we find the probability: P(X≥1)
P(X≥1) = 1 - p(x < 1)
P(X≥1) = 1 - p(x=0)
P(X≥1) = 1 - [ (e^-λ) * λ^0] / 0!
P(X≥1) = 1 - e^-2.67
P(X≥1) = 1 - 0.06945
P(X≥1) = 0.93055
P(X≥1) = 0.9306
Thus, the probability that at least 1 car arrives during the call is 0.9306.