Write c code for the following i. (Birthday paradox revisited) Compute analytica
ID: 3820933 • Letter: W
Question
Write c code for the following i. (Birthday paradox revisited) Compute analytically the probability that at least two people in a group of N persons have the same birthday Hint: Let's say, N 4. Then, the number of possible birthday combinations for 4 persons is 365 x 365 x 365 x365 The number of possible cases where 4 persons have different birthdays (no one shares the same birthday) is 365 x 365 x 365 x 365 Therefore, the probability that all 4 persons have different birthdays is 365 364 363 362 365 365 365 365 probability that at least two people out of 4 have a same birthday isExplanation / Answer
#include<stdio.h>
int main()
{
//no of people
int N, no_days_in_yr = 365,i;
float diffrent_bday_peple = 1;
float diffent_bday;
//possible_bday_combination
float possible_bday_combination(int n);
printf("Enter the number of person: ");
scanf("%d", &N);
//calculate the probality of 4 persons have different bday is as below
diffent_bday = possible_bday_combination(N) / diffrent_bday_peple;
printf("Probability of %d people have different birth day = %f ", N, diffent_bday);
printf("Prbability that at least two people out of %d have a same birth day = %.4f ", N, (1 - diffent_bday));
}
float possible_bday_combination(int N)
{
float prod = 1;
int i, no_days_in_yr= 365 ,days_in_yr =365;
//Probability that all 4 persons have different bdays is
for (i = 0; i < N; i++)
{
prod *= (float)no_days_in_yr / days_in_yr;
no_days_in_yr -= 1;
}
return prod;
}
---------------------------------------------------------------
//output1
Enter the number of person: 4
Probability of 4 people have different birth day = 0.983644
Prbability that at least two people out of 4 have a same birth day = 0.0164
//output2
Enter the number of person: 6
Probability of 6 people have different birth day = 0.959538
Prbability that at least two people out of 6 have a same birth day = 0.0405
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