# Answer to Question #33669 in C++ for Ramesh Kumar

Question #33669

An insurance company has discovered that only about 0.1% of the population is involved ina certain type of accident each year. If its 10000 policy holders were randomly selected from the population, what is the probability that not more than 5 of its clients are involved in such an accident nest year.

Expert's answer

//We use Bernoulli trials http://en.wikipedia.org/wiki/Bernoulli_trial

#include <conio.h>

#include <iostream>

using namespace std;

double _pow(double base,int n )

{

double base2 = base;

if(n == 0)

return 1;

for(int i = 1 ;i < n ;i++)

base *=base2;

return base;

}

long double C(int k , int n)

{

long double top = 1,bottom =1;

for(int i = n; i > n-k; i--)

top *= i;

for(int i = 2 ;i < k; i++)

bottom *= i;

return long double (top/bottom);

}

void main()

{

int n = 10000;

double p = 0.001 , P = 0. ;

for(int k = 0 ; k <= 5; k++)

{

P += C(k, n ) * _pow (p , k) * _pow (1 - p , n - k);

}

cout<<"Probability = "<<P*100<<"%"<<endl;

_getch();

}

#include <conio.h>

#include <iostream>

using namespace std;

double _pow(double base,int n )

{

double base2 = base;

if(n == 0)

return 1;

for(int i = 1 ;i < n ;i++)

base *=base2;

return base;

}

long double C(int k , int n)

{

long double top = 1,bottom =1;

for(int i = n; i > n-k; i--)

top *= i;

for(int i = 2 ;i < k; i++)

bottom *= i;

return long double (top/bottom);

}

void main()

{

int n = 10000;

double p = 0.001 , P = 0. ;

for(int k = 0 ; k <= 5; k++)

{

P += C(k, n ) * _pow (p , k) * _pow (1 - p , n - k);

}

cout<<"Probability = "<<P*100<<"%"<<endl;

_getch();

}

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