Ayaana Buck

2020-10-21

On average, there are 3 accidents per month at one intersection. We need to find out what is the probability that in any month at this intersection

a) What if there are 5 accidents?

b) What if there are less than 3 accidents?

c) What if there are at least 2 accidents?

berggansS

Skilled2020-10-22Added 91 answers

Given: mean $(\lambda )=3$

let "X" be the accidents per month occur at a certain intersection

$X-Poisson(\lambda =3)$

formula: $P(X=x)=\text{}\frac{{e}^{-\lambda}\text{}\cdot \text{}{\lambda}^{x}}{x}!$

a)

b)

c)

Jeffrey Jordon

Expert2021-10-07Added 2605 answers

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

$a)P[X=5]=p(5)=\frac{{e}^{-\mu}{\mu}^{x}}{x!}=\frac{{e}^{-3}{3}^{5}}{5!}=0.10082\approx 0.10$

$b)P[X<3]=F(2)=\sum _{x=0}^{2}\frac{{e}^{-\mu}{\mu}^{x}}{x!}=\sum _{x=0}^{2}\frac{{e}^{-3}{3}^{x}}{x!}=0.4232$

$c)P[X\ge 2]=1-P[X<2]=1-P[X\le 1]=1-F[1]=1-(\frac{{e}^{-3}{3}^{0}}{0!}+\frac{{e}^{-3}{3}^{1}}{1!})=0.800852\approx 0.8$

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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