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Consider the class of events. Suppose the probability that at least one of the events A or C occurs is 0.75 and the probability that at least one of the four events occurs is 0.90.Determine the probability that neither of the events A or C but at least one of the events B or D occurs.
Use the pattern and .
We use the MATLAB procedure, which displays the essential patterns.
minvec3
Variables are A, B, C, Ac, Bc, CcThey may be renamed, if desired.
E = A|~(B&C);
F = A|B|(Bc&Cc);
disp([E;F])
1 1 1 0 1 1 1 1 % Not equal1 0 1 1 1 1 1 1
G = ~(A|B);H = (Ac&C)|(Bc&C);
disp([G;H])
1 1 0 0 0 0 0 0 % Not equal0 1 0 1 0 1 0 0
K = (A&B)|(A&C)|(B&C);
disp([A;K])
0 0 0 0 1 1 1 1 % A not contained in K0 0 0 1 0 1 1 1 Use (1) minterm maps, (2) indicator functions (evaluated on minterms), (3) the m-procedure minvec3 and MATLAB logical operations to show that
We use the MATLAB procedure, which displays the essential patterns.
minvec3
Variables are A, B, C, Ac, Bc, CcThey may be renamed, if desired.
E = (A&(B|Cc))|(Ac&B&C);
F = (A&((B&C)|Cc))|(Ac&B);
disp([E;F])
0 0 0 1 1 0 1 1 % E subset of F0 0 1 1 1 0 1 1
G = A|(Ac&B&C);
H = (A&B)|(B&C)|(A&C)|(A&Bc&Cc);
disp([G;H])
0 0 0 1 1 1 1 1 % G = H0 0 0 1 1 1 1 1 Minterms for the events , arranged as on a minterm map are
0.0168 0.0072 0.0252 0.0108
0.0392 0.0168 0.0588 0.02520.0672 0.0288 0.1008 0.0432
0.1568 0.0672 0.2352 0.1008
What is the probability that three or more of the events occur on a trial? Of exactly two? Of two or fewer?
We use mintable(4) and determine positions with correct number(s) of ones (number of occurrences). An alternate is to use minvec4 and express theBoolean combinations which give the correct number(s) of ones.
npr02_04 Minterm probabilities are in pm. Use mintable(4)
a = mintable(4);s = sum(a); % Number of ones in each minterm position
P1 = (s>=3)*pm' % Select and add minterm probabilities
P1 = 0.4716P2 = (s==2)*pm'
P2 = 0.3728P3 = (s<=2)*pm'
P3 = 0.5284 Minterms for the events , arranged as on a minterm map are
0.0216 0.0324 0.0216 0.0324 0.0144 0.0216 0.0144 0.0216
0.0144 0.0216 0.0144 0.0216 0.0096 0.0144 0.0096 0.01440.0504 0.0756 0.0504 0.0756 0.0336 0.0504 0.0336 0.0504
0.0336 0.0504 0.0336 0.0504 0.0224 0.0336 0.0224 0.0336
What is the probability that three or more of the events occur on a trial? Of exactly four? Of three or fewer? Of either two or four?
We use mintable(5) and determine positions with correct number(s) of ones (number of occurrences).
npr02_05 Minterm probabilities are in pm. Use mintable(5)
a = mintable(5);s = sum(a); % Number of ones in each minterm position
P1 = (s>=3)*pm' % Select and add minterm probabilities
P1 = 0.5380P2 = (s==4)*pm'
P2 = 0.1712P3 = (s<=3)*pm'
P3 = 0.7952P4 = ((s==2)|(s==4))*pm'
P4 = 0.4784 
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