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2.3: Problems on Minterm Analysis

  • Page ID
    10866
    • Contributed by Paul Pfeiffer
    • Professor emeritus (Computational and Applied Mathematics) at Rice University

    Exercise \(\PageIndex{1}\)

    Consider the class \(\{A, B, C, D\}\) 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.

    Answer

    Use the pattern \(P(E \cup F) = P(E) + P(E^c F)\) and \((A \cup C)^c = A^c C^c\).

    \(P(A \cup C \cup B \cup D) = P(A \cup C) + P(A^c C^c (B \cup D))\), so that \(P(A^c C^c (B \cup D)) = 0.90 - 0.75 = 0.15\)

    Exercise \(\PageIndex{2}\)

    1. Use minterm maps to show which of the following statements are true for any class \(\{A, B, C\}\):

      a. \(A \cup (BC)^c = A \cup B \cup B^c C^c\)
      b. \((A \cup B)^c = A^c C \cup B^c C\)
      c. \(A \subset AB \cup AC \cup BC\)
    2. Repeat part (1) using indicator functions (evaluated on minterms).
    3. Repeat part (1) using the m-procedure minvec3 and MATLAB logical operations.
    Answer

    We use the MATLAB procedure, which displays the essential patterns.

    minvec3
    Variables are A, B, C, Ac, Bc, Cc
    They 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 equal
         1     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 equal
         0     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 K
         0     0     0     1     0     1     1     1

    Exercise \(\PageIndex{3}\)

    Use (1) minterm maps, (2) indicator functions (evaluated on minterms), (3) the m-procedure minvec3 and MATLAB logical operations to show that

    a. \(A(B \cup C^c) \cup A^c BC \subset A (BC \cup C^c) \cup A^c B\)
    b. \(A \cup A^c BC = AB \cup BC \cup AC \cup AB^c C^c\)

    Answer

    We use the MATLAB procedure, which displays the essential patterns.

    minvec3
    Variables are A, B, C, Ac, Bc, Cc
    They 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 F
         0     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 = H
         0     0     0     1     1     1     1     1

    Exercise \(\PageIndex{4}\)

    Minterms for the events \(\{A, B, C, D\}\), arranged as on a minterm map are

                         0.0168    0.0072    0.0252    0.0108
                         0.0392    0.0168    0.0588    0.0252
                         0.0672    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?

    Answer

    We use mintable(4) and determine positions with correct number(s) of ones (number of occurrences). An alternate is to use minvec4 and express the Boolean 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 =  0.4716
    P2 = (s==2)*pm'
    P2 =  0.3728
    P3 = (s<=2)*pm'
    P3 =  0.5284

    Exercise \(\PageIndex{5}\)

    Minterms for the events \(\{A, B, C, D, E\}\), 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  0.0144
           0.0504  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?

    Answer

    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 =  0.5380
    P2 = (s==4)*pm'
    P2 =  0.1712
    P3 = (s<=3)*pm'
    P3 =  0.7952
    P4 = ((s==2)|(s==4))*pm'
    P4 =  0.4784

    Exercise \(\PageIndex{6}\)

    Suppose \(P(A \cup B^c C) = 0.65\), \(P(AC) = 0.2\), \(P(A^c B) = 0.25\)

    \(P(A^c C^c) = 0.25\), \(P(BC) = 0.30\). Determine \(P((AC^c \cup A^c C) B^c)\).

    Then determine \(P((AB^c \cup A^c) C^c)\) and \(P(A^c(B \cup C^c))\), if possible.

    Answer
    % file npr02_06.m       % Data file
    % Data for Exercise 2.3.6.
    minvec3
    DV = [A|Ac; A|(Bc&C); A&C; Ac&B; Ac&Cc; B&Cc];
    DP = [1      0.65     0.20 0.25  0.25   0.30];
    TV = [((A&Cc)|(Ac&C))&Bc; ((A&Bc)|Ac)&Cc; Ac&(B|Cc)];
    disp('Call for mincalc')
    npr02_06             % Call for data
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     
     Computable target probabilities
        1.0000    0.3000     % The first and third target probability
        3.0000    0.3500     % is calculated. Check with minterm map.
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{7}\)

    Suppose \(P((AB^c \cup A^cB)C) = 0.4\), \(P(AB) = 0.2\), \(P(A^cC^c) = 0.3\), \(P(A) = 0.6\), \(P(C) = 0.5\), and \(P(AB^cC^c) = 0.1\). Determine \(P(A^c C^c \cup AC)\), \(P(AB^c \cup A^c)C^c)\), and \(P(A^c(B \cup C^c))\), if possible.

    Answer
    % file npr02_07.m
    % Data for Exercise 2.3.7.
    minvec3
    DV = [A|Ac; ((A&Bc)|(Ac&B))&C; A&B; Ac&Cc;  A;  C; A&Bc&Cc];
    DP = [ 1        0.4            0.2   0.3   0.6 0.5   0.1];
    TV = [(Ac&Cc)|(A&C); ((A&Bc)|Ac)&Cc; Ac&(B|Cc)];
    disp('Call for mincalc')
    npr02_07             % Call for data
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.7000    % All target probabilities calculable
        2.0000    0.4000    % even though not all minterms are available
        3.0000    0.4000
    The number of minterms is 8
    The number of available minterms is 6
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{8}\)

    Suppose \(P(A) = 0.6\), \(P(C) = 0.4\), \(P(AC) = 0.3\), \(P(A^cB) = 0.2\) and \(P(A^cB^cC^c) = 0.1\).

    Determine \(P((A \cup B)C^c\), \(P(AC^c \cup A^c C)\), and \(P(AC^c \cup A^cB)\), if possible.

    Answer
    % file npr02_08.m
    % Data for Exercise 2.3.8.
    minvec3
    DV = [A|Ac; A;  C;  A&C; Ac&B; Ac&Bc&Cc];
    DP = [ 1   0.6 0.4  0.3  0.2     0.1];
    TV = [(A|B)&Cc; (A&Cc)|(Ac&C); (A&Cc)|(Ac&B)];
    disp('Call for mincalc')
     
     
    npr02_08             % Call for data
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.5000    % All target probabilities calculable
        2.0000    0.4000    % even though not all minterms are available
        3.0000    0.5000
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{9}\)

    Suppose \(P(A) = 0.5\), \(P(AB) = P(AC) = 0.3\), and \(P(ABC^c) = 0.1\).

    Determine \(P(A(BC^c)^c\) and \(P(AB \cup AC \cup BC)\).

    Then repeat with additional data \(P(A^cB^cC^c) = 0.1\) and \(P(A^c BC) = 0.05\)

    Answer
    % file npr02_09.m
    % Data for Exercise 2.3.9.
    minvec3
    DV = [A|Ac;  A; A&B; A&C; A&B&Cc];
    DP = [ 1    0.5 0.3  0.3   0.1];
    TV = [A&(~(B&Cc)); (A&B)|(A&C)|(B&C)];
    disp('Call for mincalc')
     
    % Modification for part 2
    % DV = [DV; Ac&Bc&Cc; Ac&B&C];
    % DP = [DP 0.1 0.05];
    npr02_09             % Call for data
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.4000    % Only the first target probability calculable
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    DV = [DV; Ac&Bc&Cc; Ac&B&C];  % Modification of data
    DP = [DP 0.1 0.05];
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.4000             % Both target probabilities calculable
        2.0000    0.4500             % even though not all minterms are available
    The number of minterms is 8
    The number of available minterms is 6
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{10}\)

    Given \(P(A) = 0.6\), \(P(A^c B^c) = 0.2\), \(P(AC^c) = 0.4\), and \(P(ACD^c) = 0.1\).

    Determine \(P(A^c B \cup A(C^c \cup D))\).

    Answer
    % file npr02_10.m
    % Data for Exercise 2.3.10.
    minvec4
    DV = [A|Ac;  A;  Ac&Bc; A&Cc; A&C&Dc];
    DP = [1     0.6  0.2    0.4    0.1];
    TV = [(Ac&B)|(A&(Cc|D))];
    disp('Call for mincalc')
    npr02_10
    Variables are A, B, C, D, Ac, Bc, Cc, Dc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.7000             % Checks with minterm map solution
    The number of minterms is 16
    The number of available minterms is 0
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{11}\)

    A survey of a represenative group of students yields the following information:

    • 52 percent are male
    • 85 percent live on campus
    • 78 percent are male or are active in intramural sports (or both)
    • 30 percent live on campus but are not active in sports
    • 32 percent are male, live on campus, and are active in sports
    • 8 percent are male and live off campus
    • 17 percent are male students inactive in sports
    1. What is the probability that a randomly chosen student is male and lives on campus?
    2. What is the probability of a male, on campus student who is not active in sports?
    3. What is the probability of a female student active in sports?
    Answer
    % file npr02_11.m
    % Data for Exercise 2.3.11.
    % A = male;  B = on campus;  C = active in sports
    minvec3
    DV = [A|Ac;  A;   B;  A|C; B&Cc; A&B&C; A&Bc; A&Cc];
    DP = [ 1    0.52 0.85 0.78 0.30  0.32   0.08 0.17];
    TV = [A&B; A&B&Cc; Ac&C];
    disp('Call for mincalc')
     
    npr02_11
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.4400
        2.0000    0.1200
        3.0000    0.2600
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{12}\)

    A survey of 100 persons of voting age reveals that 60 are male, 30 of whom do not identify with a political party; 50 are members of a political party; 20 nonmembers of a party voted in the last election, 10 of whom are female. How many nonmembers of a political party did not vote? Suggestion Express the numbers as a fraction, and treat as probabilities.

    Answer
    % file npr02_12.m
    % Data for Exercise 2.3.12.
    % A = male;  B = party member; C = voted last election
    minvec3
    DV = [A|Ac;  A;  A&Bc;  B;  Bc&C; Ac&Bc&C];
    DP = [  1   0.60 0.30  0.50 0.20  0.10];
    TV = [Bc&Cc];
    disp('Call for mincalc')
    npr02_12
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3000
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{13}\)

    During a period of unsettled weather, let A be the event of rain in Austin, B be the event of rain in Houston, and C be the event of rain in San Antonio. Suppose:

    \(P(AB) = 0.35\), \(P(AB^c) = 0.15\), \(P(AC) = 0.20\), \(P(AB^c \cup A^cB) = 0.45\)

    \(P(BC) = 0.30\) \(P(B^c C) = 0.05\) \(P(A^c B^c C^c) = 0.15\)

    1. What is the probability of rain in all three cities?
    2. What is the probability of rain in exactly two of the three cities?
    3. What is the probability of rain in exactly one of the cities?
    Answer
    % file npr02_13.m
    % Data for Exercise 2.3.13.
    % A = rain in Austin;  B = rain in Houston;
    % C = rain in San Antonio
    minvec3
    DV = [A|Ac; A&B; A&Bc; A&C; (A&Bc)|(Ac&B); B&C; Bc&C; Ac&Bc&Cc];
    DP = [  1   0.35 0.15  0.20    0.45        0.30 0.05   0.15];
    TV = [A&B&C; (A&B&Cc)|(A&Bc&C)|(Ac&B&C); (A&Bc&Cc)|(Ac&B&Cc)|(Ac&Bc&C)];
    disp('Call for mincalc')
    npr02_13
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.2000
        2.0000    0.2500
        3.0000    0.4000
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{14}\)

    One hundred students are questioned about their course of study and plans for graduate study. Let \(A =\) the event the student is male; \(B =\) the event the student is studying engineering; \(C=\) the event the student plans at least one year of foreign language; \(D =\) the event the student is planning graduate study (including professional school). The results of the survey are:

    There are 55 men students; 23 engineering students, 10 of whom are women; 75 students will take foreign language classes, including all of the women; 26 men and 19 women plan graduate study; 13 male engineering students and 8 women engineering students plan graduate study; 20 engineering students will take a foreign language and plan graduate study; 5 non engineering students plan graduate study but no foreign language courses; 11 non engineering, women students plan foreign language study and graduate study.

    1. What is the probability of selecting a student who plans foreign language classes and graduate study?
    2. What is the probability of selecting a women engineer who does not plan graduate study?
    3. What is the probability of selecting a male student who either studies a foreign language but does not intend graduate study or will not study a foreign language but plans graduate study?
    Answer
    % file npr02_14.m
    % Data for Exercise 2.3.14.
    % A = male;  B = engineering;
    % C = foreign language; D = graduate study
    minvec4
    DV = [A|Ac; A; B; Ac&B; C; Ac&C; A&D; Ac&D; A&B&D; ...
          Ac&B&D; B&C&D; Bc&Cc&D; Ac&Bc&C&D];
    DP = [1 0.55 0.23 0.10 0.75 0.45 0.26 0.19 0.13 0.08 0.20 0.05 0.11];
    TV = [C&D; Ac&Dc; A&((C&Dc)|(Cc&D))];
    disp('Call for mincalc')
    npr02_14
    Variables are A, B, C, D, Ac, Bc, Cc, Dc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3900
        2.0000    0.2600          % Third target probability not calculable
    The number of minterms is 16
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{15}\)

    A survey of 100 students shows that: 60 are men students; 55 students live on campus, 25 of whom are women; 40 read the student newspaper regularly, 25 of whom are women; 70 consider themselves reasonably active in student affairs—50 of these live on campus; 35 of the reasonably active students read the newspaper regularly; All women who live on campus and 5 who live off campus consider themselves to be active; 10 of the on-campus women readers consider themselves active, as do 5 of the off campus women; 5 men are active, off-campus, non readers of the newspaper.

    1. How many active men are either not readers or off campus?
    2. How many inactive men are not regular readers?
    Answer
    % file npr02_15.m
    % Data for Exercise 2.3.15.
    % A = men; B = on campus; C = readers; D = active
    minvec4
    DV = [A|Ac; A;  B;  Ac&B;  C;  Ac&C;  D;  B&D; C&D; ...
         Ac&B&D; Ac&Bc&D; Ac&B&C&D; Ac&Bc&C&D; A&Bc&Cc&D];
    DP = [1  0.6 0.55 0.25 0.40 0.25 0.70 0.50 0.35 0.25 0.05 0.10 0.05 0.05];
    TV = [A&D&(Cc|Bc); A&Dc&Cc];
    disp('Call for mincalc')
    npr02_15
    Variables are A, B, C, D, Ac, Bc, Cc, Dc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3000
        2.0000    0.2500
    The number of minterms is 16
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{16}\)

    A television station runs a telephone survey to determine how many persons in its primary viewing area have watched three recent special programs, which we call a, b, and c. Of the 1000 persons surveyed, the results are:

    221 have seen at least a; 209 have seen at least b; 112 have seen at least c; 197 have seen at least two of the programs; 45 have seen all three; 62 have seen at least a and c; the number having seen at least a and b is twice as large as the number who have seen at least b and c.

    • (a) How many have seen at least one special?
    • (b) How many have seen only one special program?
    Answer
    % file npr02_16.m
    % Data for Exercise 2.3.16.
    minvec3
    DV = [A|Ac; A;    B;    C; (A&B)|(A&C)|(B&C); A&B&C; A&C; (A&B)-2*(B&C)];
    DP = [ 1  0.221 0.209 0.112   0.197           0.045  0.062      0];
    TV = [A|B|C; (A&Bc&Cc)|(Ac&B&Cc)|(Ac&Bc&C)];
    npr02_16
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3000
        2.0000    0.1030
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{17}\)

    An automobile safety inspection station found that in 1000 cars tested:

    • 100 needed wheel alignment, brake repair, and headlight adjustment
    • 325 needed at least two of these three items
    • 125 needed headlight and brake work
    • 550 needed at wheel alignment
    1. How many needed only wheel alignment?
    2. How many who do not need wheel alignment need one or none of the other items?
    Answer
    % file npr02_17.m
    % Data for Exercise 2.3.17.
    % A = alignment;  B = brake work;  C = headlight
    minvec3
    DV = [A|Ac; A&B&C; (A&B)|(A&C)|(B&C); B&C;    A  ];
    DP = [ 1    0.100      0.325          0.125 0.550];
    TV = [A&Bc&Cc; Ac&(~(B&C))];
    disp('Call for mincalc')
    npr02_17
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.2500
        2.0000    0.4250
    The number of minterms is 8
    The number of available minterms is 3
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{18}\)

    Suppose \(P(A(B \cup C)) = 0.3\), \(P(A^c) = 0.6\), and \(P(A^c B^c C^c) = 0.1\).

    Determine \(P(B \cup C)\), \(P((AB \cup A^c B^c)C^c \cup AC)\), and \(P(A^c (B \cup C^c))\), if possible.

    Repeat the problem with he additional data \(P(A^c BC) = 0.2\) and \(P(A^cB) = 0.3\).

    Answer
    % file npr02_18.m
    % Date for Exercise 2.3.18.
    minvec3
    DV = [A|Ac; A&(B|C); Ac; Ac&Bc&Cc];
    DP = [ 1     0.3     0.6    0.1];
    TV = [B|C; (((A&B)|(Ac&Bc))&Cc)|(A&C); Ac&(B|Cc)];
    disp('Call for mincalc')
    % Modification
    % DV = [DV; Ac&B&C; Ac&B];
    % DP = [DP   0.2     0.3];
    npr02_18
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.8000
        2.0000    0.4000
    The number of minterms is 8
    The number of available minterms is 2
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    DV = [DV; Ac&B&C; Ac&B];      % Modified data
    DP = [DP   0.2     0.3];
    mincalc                       % New calculation
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.8000
        2.0000    0.4000
        3.0000    0.4000
    The number of minterms is 8
    The number of available minterms is 5
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{19}\)

    A computer store sells computers, monitors, printers. A customer enters the store. Let A, B, C be the respective events the customer buys a computer, a monitor, a printer. Assume the following probabilities:

    • The probability \(P(AB)\) of buying both a computer and a monitor is 0.49.
    • The probability \(P(ABC^c)\) of buying both a computer and a monitor but not a printer is 0.17.
    • The probability \(P(AC)\) of buying both a computer and a printer is 0.45.
    • The probability \(P(BC)\) of buying both a monitor and a printer is 0.39
    • The probability \(P(AC^c \bigvee A^cC)\) of buying a computer or a printer, but not both is 0.50.
    • The probability \(P(AB^c \bigvee A^cB)\) of buying a computer or a monitor, but not both is 0.43.
    • The probability \(P(BC^c \bigvee B^c C)\) of buying a monitor or a printer, but not both is 0.43.
    1. What is the probability \(P(A)\), \(P(B)\), or \(P(C)\) of buying each?
    2. What is the probability of buying exactly two of the three items?
    3. What is the probability of buying at least two?
    4. What is the probability of buying all three?
    Answer
    % file npr02_19.m
    % Data for Exercise 2.3.19.
    % A = computer;  B = monitor;  C = printer
    minvec3
    DV = [A|Ac; A&B; A&B&Cc; A&C; B&C; (A&Cc)|(Ac&C); ...
              (A&Bc)|(Ac&B); (B&Cc)|(Bc&C)];
    DP = [1 0.49 0.17 0.45 0.39 0.50 0.43 0.43];
    TV = [A; B; C; (A&B&Cc)|(A&Bc&C)|(Ac&B&C); (A&B)|(A&C)|(B&C); A&B&C];
    disp('Call for mincalc')
    npr02_19
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.8000
        2.0000    0.6100
        3.0000    0.6000
        4.0000    0.3700
        5.0000    0.6900
        6.0000    0.3200
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{20}\)

    Data are \(P(A) = 0.232\), \(P(B) = 0.228\), \(P(ABC) = 0.045\), \(P(AC) = 0.062\), \(P(AB \cup AC \cup BC) = 0.197\) and \(P(BC0 = 2P(AC)\).

    Determine \(P(A \cup B \cup C)\) and \(P(A^c B^c C)\), if possible.

    Repeat, with the additional data \(P(C) = 0.230\).

    Answer
    % file npr02_20.m
    % Data for Exercise 2.3.20.
    minvec3
    DV = [A|Ac; A;     B;  A&B&C; A&C; (A&B)|(A&C)|(B&C); B&C - 2*(A&C)];
    DP = [  1  0.232 0.228 0.045 0.062      0.197            0];
    TV = [A|B|C; Ac&Bc&Cc];
    disp('Call for mincalc')
    % Modification
    % DV = [DV; C];
    % DP = [DP  0.230 ];
     
    npr02_20
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    mincalc
    Data vectors are linearly independent
    Data probabilities are INCONSISTENT
    The number of minterms is 8
    The number of available minterms is 6
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    disp(PMA)
        2.0000    0.0480
        3.0000   -0.0450    % Negative minterm probabilities indicate
        4.0000   -0.0100    % inconsistency of data
        5.0000    0.0170
        6.0000    0.1800
        7.0000    0.0450
    DV = [DV; C];
    DP = [DP 0.230];
    mincalc
    Data vectors are linearly independent
    Data probabilities are INCONSISTENT
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{21}\)

    Data are: \(P(A) = 0.4\), \(P(AB) = 0.3\), \(P(ABC) = 0.25\), \(P(C) = 0.65\), \(P(A^cC^c) = 0.3\). Determine available minterm probabilities and the following,

    if computable:

    \(P(AC^c \cup A^c C)\), \(P(A^cB^c)\), \(P(A \cup B)\), \(P(AB^c)\)

    With only six items of data (including \(P(\Omega) = P(A \bigvee A^c) = 1\), not all minterms are available. Try the additional data \(P(A^cB C^c) = 0.1\) and \(P(A^cB^c) = 0.3\). Are these consistent and linearly independent? Are all minterm probabilities available?

    Answer
    % file npr02_21.m
    % Data for Exercise 2.3.21.
    minvec3
    DV = [A|Ac; A;  A&B; A&B&C;  C;  Ac&Cc];
    DP = [ 1   0.4  0.3  0.25   0.65  0.3 ];
    TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];
    disp('Call for mincalc')
    % Modification
    % DV = [DV; Ac&B&Cc; Ac&Bc];
    % DP = [DP   0.1      0.3 ];
    
    npr02_21
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3500
        4.0000    0.1000
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    DV = [DV; Ac&B&Cc; Ac&Bc];
    DP = [DP   0.1      0.3 ];
    mincalc
    Data vectors are linearly independent
     Computable target probabilities
        1.0000    0.3500
        2.0000    0.3000
        3.0000    0.7000
        4.0000    0.1000
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA

    Exercise \(\PageIndex{22}\)

    Repeat Exercise with \(P(AB)\) changed from 0.3 to 0.5. What is the result? Explain the reason for this result.

    Answer
    % file npr02_22.m
    % Data for Exercise 2.3.22.
    minvec3
    DV = [A|Ac; A;  A&B; A&B&C;  C;  Ac&Cc];
    DP = [ 1   0.4  0.5  0.25   0.65  0.3 ];
    TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];
    disp('Call for mincalc')
    % Modification
    % DV = [DV; Ac&B&Cc; Ac&Bc];
    % DP = [DP   0.1      0.3 ];
    
    npr02_22
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are linearly independent
    Data probabilities are INCONSISTENT
    The number of minterms is 8
    The number of available minterms is 4
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    disp(PMA)
        4.0000   -0.2000
        5.0000    0.1000
        6.0000    0.2500
        7.0000    0.2500
    DV = [DV; Ac&B&Cc; Ac&Bc];
    DP = [DP   0.1      0.3 ];
    mincalc
    Data vectors are linearly independent
    Data probabilities are INCONSISTENT
    The number of minterms is 8
    The number of available minterms is 8
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    disp(PMA)
             0    0.2000
        1.0000    0.1000
        2.0000    0.1000
        3.0000    0.2000
        4.0000   -0.2000
        5.0000    0.1000
        6.0000    0.2500
        7.0000    0.2500

    Exercise \(\PageIndex{23}\)

    Repeat Exercise with the original data probability matrix, but with \(AB\) replaced by \(AC\) in the data vector matrix. What is the result? Does mincalc work in this case? Check results on a minterm map.

    Answer
    % file npr02_23.m
    % Data for Exercise 2.3.23.
    minvec3
    DV = [A|Ac; A;  A&C; A&B&C;  C;  Ac&Cc];
    DP = [ 1   0.4  0.3  0.25   0.65  0.3 ];
    TV = [(A&Cc)|(Ac&C); Ac&Bc; A|B; A&Bc];
    disp('Call for mincalc')
    % Modification
    % DV = [DV; Ac&B&Cc; Ac&Bc];
    % DP = [DP   0.1      0.3 ];
    npr02_23
    Variables are A, B, C, Ac, Bc, Cc
    They may be renamed, if desired.
    Call for mincalc
    mincalc
    Data vectors are NOT linearly independent
    Warning: Rank deficient, rank = 5  tol =    5.0243e-15
     Computable target probabilities
        1.0000    0.4500
    The number of minterms is 8
    The number of available minterms is 2
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
    DV = [DV; Ac&B&Cc; Ac&Bc];
    DP = [DP   0.1      0.3 ];
    mincalc
    Data vectors are NOT linearly independent
    Warning: Matrix is singular to working precision.
     Computable target probabilities
         1   Inf             % Note that p(4) and p(7) are given in data
         2   Inf
         3   Inf
    The number of minterms is 8
    The number of available minterms is 6
    Available minterm probabilities are in vector pma
    To view available minterm probabilities, call for PMA
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