/************************************************************
* Evaluate probabilities by parsing a string expression *
* and performing the implied calculations *
* --------------------------------------------------------- *
* SAMPLE RUN: *
* *
* Type expression to be evaluated (no syntax checking). *
* Example: '1-c(10,0,0.2)-c(10,1,0.2)' *
* *
* 'c(10,0,.1)+c(10,1,0.1)+c(10,2,0.1)' *
* *
* C(10,0) = 1 *
* C(10,1) = 10 *
* C(10,2) = 45 *
* probability = 0.9298 *
* *
* --------------------------------------------------------- *
* Ref.: "Problem Solving with Fortran 90 By David R.Brooks, *
* Springer-Verlag New York, 1997". *
* *
* C++ Release By J-P Moreau, Paris. *
* (www.jpmoreau.fr) *
*************************************************************
!Explanations:
!------------
! A manufacturer's experience has shown that 10% of all integrated
! circuits (ICs) will be defective. Because of this high failure rate,
! a quality control engineer monitors the manufacturing process by
! testing random samples every day. What is the probability that:
! (a) exactly two ICs in a sample of 10 will be defective?
! (b) at least two will be defective?
! (c) no more than two will be defective?
! Answers:
! (a) the probability that a particular sample of 10 will contain
! 2 8
! exactly two defective ICs is: (0.1) (0.9) = 0.004305. However,
! there are C(10,2)=45 possible combinations of two defective and eight
! good ICs. From probability theory, the number of combinations of n
! things taken k at a time is: C(n,k) = n!/[k!(n-k)!] where
! ! indicates the factorial function (n! = 1 x 2 x3 x ....n). Therefore,
! the probability that a sample of 10 will contain exactly two defects
! is: 2 8
! P(=2) = C(10,2)(0.1) (0.9) = 45 x 0.004305 = 0.1937
!
! (b) the probability of finding at least two defective ICs is equal to
! 1 minus the probability of 0 defective IC minus the probability of
! 1 defective IC: 0 10 1 9
! P(>=2) = 1 - C(10,0)(0.1) (0.9) - C(10,0)(0.1) (0.9) = 0.2639
!
! (Reemeber that 0! = 1 by definition.)
!
! (c) the probability of finding no more than two defective ICs is:
! 0 10 1 9
! P(<=2) = C(10,0)(0.1) (0.9) + C(10,1)(0.1) (0.9)
! 2 8
! + C(10,2)(0.1) (0.9) = 0.9298
!
! For example, for part (b) of the problem, the user will type the
! string: '1-c(10,0,.1)-c(10,1,.1)'.
!---------------------------------------------------------------------*/
#include
#include
#include
#include
char a[80]; // string to be calculated
char s[20], s1[20];
int i, j, jj, k, len, n, sign, left;
double probability, prob_a;
double Fact(double x) {
// Calculate x!
double prod;
int i,ix;
prod = 1.0;
ix=(int) floor(x);
for (i=2; i<=ix; i++) prod *= i;
return (prod);
}
int C(int n, int k) {
// Calculate combinations of n things taken k at a time
double temp, denom; int ic;
denom = Fact(1.0*k)*Fact(1.0*(n-k));
temp = Fact(n) / denom;
ic=(int) floor(temp);
return (ic);
}
void main() {
printf("\n Type expression to be evaluated (no syntax checking).\n");
printf(" Example: 1-c(10,0,0.2)-c(10,1,0.2)\n\n");
printf(" "); scanf("%s", a);
printf("\n");
len = strlen(a);
probability = 0.0;
if (a[0]=='1') probability = 1.0;
sign=1; // a leading + sign is optional
for (i=0; i