On average, 3 traffic accidents per month occur at a certain intersection. What is the probability that in any given month at this intersection (a) exactly 5 accidents will occur? (b) fewer than 3 accidents will occur? (c) at least 2 accidents will occur?

Ayaana Buck 2020-10-21 Answered
On average, 3 traffic accidents per month occur at a certain intersection. What is the probability that in any given month at this intersection
(a) exactly 5 accidents will occur?
(b) fewer than 3 accidents will occur?
(c) at least 2 accidents will occur?
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berggansS
Answered 2020-10-22 Author has 91 answers

Given: mean (λ)=3
let "X" be the accidents per month occur at a certain intersection
XPoisson(λ=3)
formula: P(X=x)= eλ  λxx!
a)image
b)image
c)image

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Jeffrey Jordon
Answered 2021-10-07 Author has 2047 answers

We will assume that the number of accidents at the intersection follows a Poisson distribution with a mean (\mu) of 3 per month

a)P[X=5]=p(5)=eμμxx!=e3355!=0.100820.10

b)P[X<3]=F(2)=x=02eμμxx!=x=02e33xx!=0.4232

c)P[X2]=1P[X<2]=1P[X1]=1F[1]=1(e3300!+e3311!)=0.8008520.8

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Answered 2022-03-02

a) 

λ=3, t=1, e=2.71828, X=5

=(e^(-3))*(3^(5)) ÷ 5!

=12.0982576134 ÷ 120

=0.10081881344 

 

b)

X=0,1,2 as X<3

p(0;3)+p(1;3)+p(2;3)

p(0;3) = (e^(-3))*(3^(0)) ÷ 0! = 0.04978706836 / 1 = 0.04978706836

p(1;3) = (e^(-3))*(3^(1)) ÷ 1! = 0.1493612051 

p(2;3) = (e^(-3))*(3^(2)) ÷ 2! = 0.44808361531 / 2 = 0.22404180765 

p(0;3) + p(1;3) + p(2;3) = 0.04978706836 + 0.1493612051 + 0.22404180765 = 0.42319008111 

 

c)

X>=2

[1- ( p(0;3) + p(1;3) + p(2;3) ) ] 

As I already derived the values for p(0;3), p(1;3) & p(2;3)

I can directly solve it :

1 - 0.42319008111 = 0.57680991889

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