Assuming that the chance of a traffic accident in a day in a street of Delhi is 0 001 - Maths - Probability

Question

Assuming that the chance of a traffic accident in a day in a
street of Delhi is 0 001
On how many days, out of a total of 500 days can we expect (I)
no accident (ii) more than 3 accidents, if there are 1000
such streets in the whole city ( use 1/e = 0 3679)??

Vipin Verma · Accepted Answer

It is question of poisson's distribution
Let x be the number of streets in which accident occurs be the
poisson variable
By poisson's distributions
P(x=r) =e-mmrr!, r = 0,1,2 Here n =1000, p =0 001 therefore m =np
=1000× 001=1Hence P(x=r) =e-1(1)rr!=e-1r!, r = 0,1,2 1000(1)
P(accident in no street)=P(x=0) =e-10!=0 36791=0 3679Hence no of
days expecting accident in no street =N×P(x=0) = 500×0
3679 =184 Days(approx)(2)P(Accident in more than three
street)=P(x>3)=1-P(x≤3)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)]=1-[e-10!+e-11!+e-12!+e-13!]=1-0
3679[83]=0 0189Hence no of days expecting accident in more than
three streets =N×P(x>3) =500×0 0189 =9days(approx)

Assuming that the chance of a traffic accident in a day in a street of Delhi is 0.001. On how many days, out of a total of 500 days can we expect (I) no accident (ii) more than 3 accidents, if there are 1000 such streets in the whole city. ( use 1/e = 0.3679)??

Assuming that the chance of a traffic accident in a day in a street of Delhi is 0.001.

On how many days, out of a total of 500 days can we expect (I) no accident (ii) more than 3 accidents, if there are 1000

such streets in the whole city. ( use 1/e = 0.3679)??