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Quantitative Methods Session 15 – 16.08.13 SET THEORY Pranjoy Arup Das

Quantitative Methods Session 15 – 16.08.13 SET THEORY Pranjoy Arup Das. A SET is a collection of well defined, specific objects. Each object in a SET is termed as an element. Eg . The collection of the vowels – a,e,i,o,u – is a SET

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Quantitative Methods Session 15 – 16.08.13 SET THEORY Pranjoy Arup Das

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  1. Quantitative MethodsSession 15 – 16.08.13SET THEORYPranjoy Arup Das

  2. A SET is a collection of well defined, specific objects. Each object in a SET is termed as an element. • Eg. The collection of the vowels – a,e,i,o,u – is a SET The collection of all states in the Indian Union – is a SET The collection of batsmen who have scored more than 10 century’s – a SET The collection of good cricketers – NOT A SET , because it is not specific and not well defined. • Consider the following cases: A) In a group, 2 persons can speak hindi, 2 persons can speak english 1 person can speak both hindi and english. How many people are there in the group? B) In a group, 2 persons can speak hindi out of which 1 can speak only hindi 2 persons can speak english out of which 1 can speak only english 1 person speaks both hindi and english. How many people are there in the group?

  3. C) In a group, 2 persons can speak hindi out of which 1 can speak only hindi 2 persons can speak english out of which 1 can speak only english 1 person speaks both hindi and english. 1 speaks neither hindi nor english. Eg. 1) In a group of 3 people, 2 persons can speak hindi 2 persons can speak english How many can speak both? NOTE: When there are two SET’s, say SET A and SET B with common elements • Total no. of elements of SET A and SET B= All elements of SET A + (All elements of SET B – Common elements of both A & B) In short : Total of A & B = All of A + All of B – Common of A & B Also, • All elements of SET A = Elements of only SET A + common elements of A & B • All elements of SET B = Elements of only SET B + common elements of A & B So In Eg. 1, No. of persons who speak both = 2 (all hindi)+2 (all english) – 3 (total) = 1 person.

  4. This concept can be shown with the help of diagrams called Venn diagrams. ( Concept of Venn diagrams was proposed by Swiss mathematician Euler. Later on British mathematician John Venn brought the concept into practice . So Venn diagrams are sometimes called Venn - Euler diagrams) UNIVERSAL SET ELEMENTS NEITHER OF SET A NOR OF SET B ELEMENTS OF SET A ELEMENTS OF SET B SET A SET B

  5. Can be broken into three parts SET A SET B Both SET A & SET B Both SET A & SET B = Only SET A + + Only SET B - Both SET A & SET B All of SET B All of SET A + Total no. of elements = of Set A & Set B together OR - Both SET A & SET B All of SET B All of SET A +

  6. Eg 2) In a group of 4 people, 2 persons can speak Hindi 2 persons can speak English 1 person speaks both Hindi and English. This means there are people in the group who speak neither Hindi nor English In such a case: the relationship will be - Total no. of elements = All elements of SET A – common elements of both A &B + All elements of SET B + Neither SET A nor SET B In short , TOTAL OF A & B = ALL OF A – COMMON OF A &B + ALL OF B + NEITHER A NOR B So in Eg 2, No. of persons who speak neither: (TOTAL) – (SET A) + (Common of A &B) – (SET B) = Neither A nor B 4 – 2 + 1 – 2 = 1

  7. Solved example 1) In a club, there are 35 members. 25 of them are learning German and 18 members are learning French. How many are learning both French and German? Solution: Total no. of elements = 35 SET A = members learning German SET B= members learning French. Total elements of SET A = 25 Total elements of SET B = 18 We have to find out the common elements of SET A and SET B We know that Total of A & B = All of A + All of B – Common of A & B So in this case, Members learning both = 25 + 18 – 35 = 8

  8. Solved example 2) In a survey of 700 students in a college, 180 were found to prefer Sprite, 275 were found to prefer 7Up and 95 were found to prefer both. How many students prefer neither Sprite nor 7Up? Solution: Total no. of elements = 700 SET A = Students preferring Sprite SET B= Students preferring 7Up. Total elements of SET A = 180 Total elements of SET B = 275 Common elements of SET A and SET B = 95 We have to find out the number of elements that belong to neither SET A nor B We know that TOTAL OF A & B = ALL OF A – COMMON OF A &B + ALL OF B + NEITHER A NOR B NEITHER A NOR B = TOTAL OF A & B – ALL OF A + COMMON OF A & B – ALL OF B So in this case, Members preferring neither = 700 – (180-95 ) - 275 = ___________

  9. Solved example 3) In a class of 35 students, there is an option to chose between mathematics and economics or take both. 17 have opted for mathematics, 10 have opted for mathematics but not economics. Find : • The no. of students who have taken both mathematics and economics • The no. of students who have taken economics but not mathematics. Solution: Total no. of elements = 35 SET A = Students opting mathematics SET B= Students opting for economics Total elements of SET A = 17 Elements of only SET A = 10 Total elements of SET B = ? We have to find out Elements of only SET B & Common elements of SET A & B We know that – • All elements of SET A = Elements of only SET A + common elements of A & B => 17 = 10 + Common elements => Common elements = 17 – 10 = 7 • Total of A & B = All of A– Common of A & B + All of B Find All elements of SET B and then find out Elements of only SET B.

  10. SET B SET A Problems with three types of SET s: • When there are three types of SET s, say SET A, SET B & SET C, it is best to take the help of Venn diagrams. SET A & B But not C Only SET A Only SET B All 3 Types A, B & C SET B &C But not A SET A & C But not B Only SET C SET C

  11. So Total no. of Elements = Only SET A + Only SET B + Only SET C + SET A &B not C + SET B &C not A + SET C & A not B + All 3 types A, B & C + Elements of neither A nor B nor C For the sake of convenience let us assign alphabets to each section: So, Total no. of elements – Elements of neither A, B nor C = a + b + c + d + e + f + g SET B SET A e b a d g f c SET C

  12. SET B SET A e b a d g f c SET C All elements of SET A = a + e + d + g All elements of SET B = b + f + d + e All elements of SET C = c + g + d + f Total no. of elements – Elements of neither A, B nor C = a + b + c + d + e + f + g

  13. Solved example 4) In a survey of 25 students, it was found that15 had taken mathematics, 12 had taken physics and 11 had taken chemistry. 5 had taken mathematics and chemistry, 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all three subjects. We need to find the no of students who had: a) Only chemistry b) Only mathematics c) Only physics d) Physics and chemistry but not mathematics e) Mathematics and physics but not chemistry f) Only one of the subjects g) None of the subjects

  14. Given that : In a survey of 25 students, it was found that15 had taken mathematics, 12 had taken physics and 11 had taken chemistry. 5 had taken mathematics and chemistry, 9 had taken mathematics and physics, 4 had taken physics and chemistry and 3 had taken all three subjects. Formulation: Total no. of elements = 25 SET A = Students opting mathematics, Say SET M SET B= Students opting for physics, Say SET P SET C = students opting for chemistry, Say SET C Total elements of M = 15 Total elements of P = 12 Total elements of C = 11 Both M & P = 9 Both M & C = 5 Both P & C = 4 All three types = 3 Let us represent this data on a Venn diagram.

  15. P M e =6 b = 2 a = 4 d = 3 g f = 1 = 2 c = 5 C Common Elements of M, P & C = d = 3 Common Elements of M & P = e + d = 9 => e = 9 -3 = 6 Common elements of M & C = g + d =5 => g = 5 -3 = 2 Common elements of P & C = f + d = 4 => f = 4 -3 = 1 All elements of M = a + e + d + g = 15 => a = 15 – (6 + 3 + 2) = 4 All elements of P = b + f + d + e = 12 => b = 12 – (1 + 3 + 6) = 2 All elements of C = c + g + d + f = 11 => c = 11 – (2 + 3 + 1) = 5

  16. P M e =6 a = 4 b = 2 d = 3 We need to find the no of students who had: a) Only chemistry Solution: That is, the value of c = b) Only mathematics Value of c) Only physics Value of d) Physics and chemistry but not mathematics Value of e) Mathematics and physics but not chemistry Value of f) Only one of the subjects Value of g) None of the subjects Value of 25 – (4 + 2+5+3+6+1+2) = _______ a= 4 g f = 1 = 2 b= 2 c = 5 f= 1 C e= 6 a + b + c = 4+2+5=11 Total no. of elements – ( a + b + c + d + e + f + g) =

  17. END OF CHAPTER

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