DECLARE FUNCTION Fact! (x!) DECLARE FUNCTION IC% (n%, k%) '************************************************************* '* 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". * '* * '* Basic 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)'. '----------------------------------------------------------------------- 'Program Prob DEFINT I-N DIM a AS STRING 'string to be calculated DIM s AS STRING DIM s1 AS STRING ' i, j, k, length, n, isign, left: Integer ' probability, proba: Real CLS PRINT PRINT " Type expression to be evaluated (no syntax checking)." PRINT " Example: 1-c(10,0,0.2)-c(10,1,0.2)" PRINT INPUT " ", a PRINT length = LEN(a) probability = 0! IF MID\$(a, 1, 1) = "1" THEN probability = 1! isign = 1 'a leading + sign is optional FOR i = 1 TO length IF MID\$(a, i, 1) = "+" THEN isign = 1 IF MID\$(a, i, 1) = "-" THEN isign = -1 IF MID\$(a, i, 1) = "(" THEN left = i IF MID\$(a, i, 1) = ")" THEN 'Read(a(left+1:i-1),*) n, k, prob_a s = "" FOR j = left + 1 TO i s = s + MID\$(a, j, 1) NEXT j s1 = "": j = 1 WHILE MID\$(s, j, 1) <> "," AND MID\$(s, j, 1) <> ")" s1 = s1 + MID\$(s, j, 1): j = j + 1 WEND n = VAL(s1) s1 = "": j = j + 1 WHILE MID\$(s, j, 1) <> "," AND MID\$(s, j, 1) <> ")" s1 = s1 + MID\$(s, j, 1): j = j + 1 WEND k = VAL(s1) s1 = "": j = j + 1 WHILE MID\$(s, j, 1) <> "," AND MID\$(s, j, 1) <> ")" s1 = s1 + MID\$(s, j, 1): j = j + 1 WEND PRINT proba = VAL(s1) probability = probability + isign * IC(n, k) * proba ^ k * (1! - proba) ^ (n - k) PRINT " C("; n; ","; k; ") = "; IC(n, k) END IF NEXT i F\$ = "##.####" PRINT " probability = "; PRINT USING F\$; probability END 'of main program FUNCTION Fact (x) ' Calculate x! prod = 1! ix = INT(x) FOR i = 2 TO ix prod = prod * i NEXT i Fact = prod END FUNCTION FUNCTION IC (n, k) ' Calculate combinations of n things taken k at a time denom = Fact(1! * k) * Fact(1! * (n - k)) temp = Fact(1! * n) / denom IC = INT(temp) END FUNCTION