Minterms  (Page 4/4)

Survey on personal computers

A survey of 1000 students shows that 565 have PC compatible desktop computers, 515 have Macintosh desktop computers, and 151 have laptop computers. 51 have all three,124 have both PC and laptop computers, 212 have at least two of the three, and twice as many own both PC and laptop as those who have both Macintosh desktop and laptop.A person is selected at random from this population. What is the probability he or she has at least one of these types of computer? Whatis the probability the person selected has only a laptop?

SOLUTION

Let $A=$ the event of owning a PC desktop, $B=$ the event of owning a Macintosh desktop, and $C=$ the event of owning a laptop. We utilize a minterm map for three variables to help determine minterm patterns. For example,the event $AC=M_5\bigvee M_7$ so that $P\left(AC\right)=p\left(5\right)+p\left(7\right)=p(5,7)$ .

The data, expressed in terms of minterm probabilities, are:

$P\left(A\right)=p(4,5,6,7)=0.565$ , hence $P\left(A^c\right)=p(0,1,2,3)=0.435$

$P\left(B\right)=p(2,3,6,7)=0.515$ , hence $P\left(B^c\right)=p(0,1,4,5)=0.485$

$P\left(C\right)=p(1,3,5,7)=0.151$ , hence $P\left(C^c\right)=p(0,2,4,6)=0.849$

$P\left(ABC\right)=p\left(7\right)=0.051P\left(AC\right)=p(5,7)=0.124$

$P(AB\cup AC\cup BC)=p(3,5,6,7)=0.212$

$P\left(AC\right)=p(5,7)=2p(3,7)=2P\left(BC\right)$

We use the patterns displayed in the minterm map to aid in an algebraic solution for the various minterm probabilities.

$p\left(5\right)=p(5,7)-p\left(7\right)=0.124-0.051=0.073$

$p(1,3)=P\left(A^{c}C\right)=0.151-0.124=0.027P\left(A{C}^c\right)=p(4,6)=0.565-0.124=0.441$

$p(3,7)=P\left(BC\right)=0.124/2=0.062$

$p\left(3\right)=0.062-0.051=0.011$

$p\left(6\right)=p(3,4,6,7)-p\left(3\right)-p(5,7)=0.212-0.011-0.124=0.077$

$p\left(4\right)=P\left(A\right)-p\left(6\right)-p(5,7)=0.565-0.077-0.1124=0.364$

$p\left(1\right)=p(1,3)-p\left(3\right)=0.027-0.11=0.016$

$p\left(2\right)=P\left(B\right)-p(3,7)-p\left(6\right)=0.515-0.062-0.077=0.376$

$p\left(0\right)=P\left(C^c\right)-p(4,6)-p\left(2\right)=0.849-0.441-0.376=0.032$

We have determined the minterm probabilities, which are displayed on the minterm map [link] . We may now compute the probability of any Boolean combination of the generating events $A,B,C$ . Thus,

Opinion survey

A survey of 1000 persons is made to determine their opinions onfour propositions. Let $A,B,C,D$ be the events a person selected agrees with the respective propositions. Survey results show the following probabilities forvarious combinations:

Determine the probabilities for each minterm and for each of the following combinations

$A^c(B{C}^c\cup B^{c}C)$ – that is, not A and ( B or C , but not both)

$A\cup B{C}^c$ – that is, A or ( B and not C )

SOLUTION

At the outset, it is not clear that the data are consistent or sufficient to determine the minterm probabilities. However, an examination of the data shows that there are sixteen items (includingthe fact that the sum of all minterm probabilities is one). Thus, there is hope, but no assurance, that a solution exists. A step elimination procedure, as in the previous examples,shows that all minterms can in fact be calculated. The results are displayed on the minterm map in [link] . It would be desirable to be able to analyze the problem systematically. The formulation above suggests a more systematic algebraic formulation which should makepossible machine aided solution.

The software survey problem reformulated

The data, expressed in terms of minterm probabilities, are:

$P\left(A\right)=p(4,5,6,7)=0.80$

$P\left(B\right)=p(2,3,6,7)=0.65$

$P\left(C\right)=p(1,3,5,7)=0.30$

$P\left(ABC\right)=p\left(7\right)=0.10$

$P\left(A^c{B}^c\right)=p(0,1)=0.05$

$P(AB\cup AC\cup BC)=p(3,5,6,7)=0.65$

$P\left(A{B}^{c}C\right)=p\left(5\right)=2p\left(3\right)=2P\left(A^{c}BC\right)$ , so that $p\left(5\right)-2p\left(3\right)=0$

We also have in any case

$P\left(\Omega \right)=P(A\cup A^c)=p(0,1,2,3,4,5,6,7)=1$

to complete the eight items of data needed for determining all eight minterm probabilities. The first datum can be expressed as an equation in mintermprobabilities:

This is an algebraic equation in $p\left(0\right),\cdots ,p\left(7\right)$ with a matrix of coefficients

The others may be written out accordingly, giving eight linear algebraic equations in eight variables $p\left(0\right)$ through $p\left(7\right)$ . Each equation has a matrix or vector of zero-one coefficients indicating which minterms are included. These may be written in matrix form as follows:

• The patterns in the coefficient matrix are determined by logical operations . We obtained these with the aid of a minterm map.
• The solution utilizes an algebraic procedure , which could be carried out in a variety of ways, including several standard computer packages for matrix operations.

We show in the module Minterm Vectors and MATLAB how we may use MATLAB for both aspects.