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An insurance company insures 4000 people against loss of both eyes in a car accident. Based on previous data, the rates were computed on the assumption that on the average10 persons in100000 will have car accident each year that result in this type of injury. What is the probability that more than 3 of the insured will collect on their policy in a given year?

An insurance company insures 4000
people against loss of both eyes in a car
accident. Based on previous data, the rates were computed on the assumption
that on the average10
persons in100000
will have car accident each year
that result in this type of injury. What is the probability that more than 3 of the
insured will collect on their policy in a given year?

Grade:12th pass

3 Answers

Md Mushahid
21 Points
4 years ago
Number of deaths from a disease.
No. of suicides reported in a city.
No. of printing mistakes at each page of a book.
Emission of radioactive particles.
No. of telephone calls per minute at a small business.
No. of paint spots per new automobile.
No. of arrivals at a toll booth
No. of flaws per bolt of cloth
Rahul kumar
13 Points
3 years ago
use possion distribution 
 
here probability of person to get injuery of such kind p = 10/100000
n = 4000
mean = m =  np = 4000X(10/100000) = 0.4
 
p(x) = e-m.mx / x!
 
put x = 0,1,2 
 
then add all the values and subtract it from 1 
 
that’s your answer
DEVAVRAT SINGH
26 Points
2 years ago
We can use formula of 
P(r>=n)=1-p(r
So for more than 2 persons
P(r>=2)=1-p(r
            =1-p(0)+p(1)
          since m=0.4
          So,p(r)=(e^-m*m^r)/r!
      P(r>=2)=0.007926

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