The number of these 120 intervals in which exactly three cars arrive is; 21 intervals
We are told that that no cars arrive in 18 of these 120 intervals. Thus; P(n) = 18/120
P(n) = 0.15
Using the Poisson process, we have the formula;
P(n) = [tex]\frac{(\lambda t)^{n} * e^{-\lambda t}}{n!}[/tex]
We already have that; P(n = 0) = 0.15
Thus; using t = 20 seconds, we have;
0.15 = [tex]\frac{(\lambda * 20)^{0} * e^{-20\lambda}}{0!}[/tex]
⇒ 0.15 = [tex]e^{-20\lambda}}[/tex]
⇒ In 0.15 = -20λ
⇒ -1.897 = -20λ
λ = -1.897/-20
λ = 0.09485
Thus, for exactly 3 cars;
P(n = 3) = [tex]\frac{(0.09485 * 20)^{3} * e^{-20 * 0.09485}}{3!}[/tex]
P(n = 3) = 0.1707
Thus;
Number of intervals in which exactly three cars arrive is;
120 * 0.1707 ≈ 21 intervals
Read more about Poisson Process at; https://brainly.com/question/10196617
