Supplementary Exercise 4.60 of IPS7e
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Students A, B and C are selected randomly from a population. Each of them
have probability 0.4 of opposing funding; the corresponding events are
denoted A-, B- and C-. The complementary events (denoted A+, B+ and C+) 
of supporting funding have probability 1-0.4 = 0.6. The events for
different students are assumed independent.

(a) P(A+ and B+ and C-) = P(A+)*P(B+)*P(C-) = 0.6*0.6*0.4 = 0.144,
by the multiplication rule.

(b) Possible outcomes and probabilities:
P(A+ and B+ and C+) = 0.6*0.6*0.6 = 0.216
P(A- and B+ and C+) = 0.4*0.6*0.6 = 0.144
P(A+ and B- and C+) = 0.6*0.4*0.6 = 0.144
P(A+ and B+ and C-) = 0.6*0.6*0.4 = 0.144
P(A+ and B- and C-) = 0.6*0.4*0.4 = 0.096
P(A- and B+ and C-) = 0.4*0.6*0.4 = 0.096
P(A- and B- and C+) = 0.4*0.4*0.6 = 0.096
P(A- and B- and C-) = 0.4*0.4*0.4 = 0.064

(c) X = number of students opposing funding
P(X=0) = P(A+ and B+ and C+) = 0.216
P(X=1) = P(A- and B+ and C+) + P(A+ and B- and C+) + P(A+ and B+ and C-)
       = 0.144*3 = 0.432
P(X=2) = 0.096*3 = 0.288
P(X=3) = 0.064

X follows a binomial distribution B(3,0.4), because the situation
described meets all conditions of a binomial setting.

(d) X>=2 expresses that a majority of students oppose funding
P(X>=2) = P(X=2) + P(X=3) = 0.352

Note: Because X follows a binomial distribution B(3,0.4), the probabilities 
of (c) and (d) can also be found in a binomial distribution table (e.g. Table 1
in Stephens, or the table below generated in Minitab).


Minitab commands to display the distribution of X:

 Name c1 "x"
 Set 'x'
  1( 0 : 3 / 1 )1
  End.
 Name c2 "p(x)"
 PDF 'x' 'p(x)';
  Binomial 3 0.4.
 Print 'x' 'p(x)'.

Data Display 

Row  x   p(x)
  1  0  0.216
  2  1  0.432
  3  2  0.288
  4  3  0.064

 DPlot;
  Distribution;
    Binomial 3 .4.