Paper 4: Fundamentals of Business Mathematics & Statistic

2.12 I FUNDAMENTALS OF BUSINESS MATHEMATICS AND STATISTICS

Algebra

In a finite set, if operations are made, some new subsets will be formed. In this section we will find the values

of these new subjects. Since A is a finite set, we shall denote it by n (A) for the finite elements in A, which may

be obtained by actual counting. But for unions of two or more sets, we have different formulae :

1. For union of Two sets :
   For two sets A and B which are not disjoint.
   n (A ∪ B) = n (A) + n (B) – (A ∩ B)


2. For Union of Three Sets :
   Let A, B and C be the three sets (no mutually disjoint); then
   n (A ∪ B ∪ C)= n (A) + n (B) + n (C) –n (A ∩ B) –n(C ∩ A) –n(B ∩ C)+ n (A ∩ B ∩ C)
   Note : (i) n (A ∩ B’) = n (A)–n(A ∩ B) (ii) n (A ∩ B)’ = (A’ ∪ B’)
   SOLVED EXAMPLES
   Example 13 : In a class of 100 students, 45 students read Physics, 52 students read Chemistry and 15 students
   read both the subjects. Find the number of students who study neither Physics nor Chemistry.
   Solution:
   We know n (A ∪ B) = n (A) + n (B) n (A ∩ B). Let A indicates Physics, B for Chemistry. Now
   n (A) = 45, n (B) = 52, n(A ∩ B) = 15
   So, n (A ∪ B) = 45 + 52– 15 = 82.
   We are to find n (A ́ ∩ B ́) = n (A ∪ B) ́ = 100– n(A ∪ B) = 100– 82 = 18.
   Example 14 : In a class of 30 students, 15 students have taken English, 10 Students have taken English but
   not French. Find the number of students who have taken (i) French and(ii) French but not English.
   Solution:
   Let E stands for English, F for French.
   n ( E ∪ F) = 30, n (E) = 15, n (E ∩ F ́) = 10
   n (E ∪ F) = n (E) + n ( F) –n (E ∩ F) ....(i)
   Now n (E ∩ F ́) = n (E) –n (E ∩ F)
   or, 10 = 15 –n (E ∩ F), or, n (E ∩ F) = 15 –10 = 5
   From (i) , 30 = 15 + n (F) –5 or, n(F) = 20
   n (F ∩ E’) = n (F) –n(F ∩ E) = 20 –5 = 15.
   Example 15 : In a class of 50 students, 15 read Physics, 20 Chemistry and 20 read Mathematics, 3 read
   Physics and Chemistry, 6 read Chemistry and Mathematics and 5 read Physics and Mathematics, 7 read
   none of the subjects. How many students read all the three subjects?
   Solution:
   Let A stands for Physics, B for Chemistry, C for Mathematics
   Now n (A) = 15, n (B) = 20, n (C) = 20
   n (A ∩ B) = 3, n (B ∩ C) = 6, n (C ∩ A) = 5, n (A ∩ B ∩ C) =?
   and n (A ∪ B ∪ C) = 50 – 7 = 43, as 7 students read nothing.
   From n (A∪B∪C)
   = n (A) + n (B) + n (C) – n (A ∩ B) – n (B ∩ C) – n ( C ∩ A) + n (A ∩ B ∩ C)