A college awarded 38 medals in football 15 in basketball 20 cricket if these medals went to a picture of 58 men and men got medals in all the three sports. Find number of medals received by the men exactly in two of the three sports.
Answers
Answer:
Let F,B,C be the sets of students who received medals in football, basketball, and cricket respectively.
n(F)=38
n(B)=15
n(C)=20
n(F∪B∪C)=58
n(F∩B∩C)=3
We know, n(F∪B∪C)=n(F)+n(B)+n(C)−n(F∩B)−n(B∩C)−n(F∩C)+n(F∩B∩C)
n(F∩B)+n(B∩C)+n(C∩F)=73−58+3=18
the number of students who received medals in exactly two of the three
sports is,
n(F∩B)+n(B∩C)+n(C∩F)−3×n(F∩B∩C)=18−9=9
Step-by-step explanation:
Let F,B,C be the sets of students who received medals in football, basketball, and cricket respectively.
n(F)=38
n(B)=15
n(C)=20
n(F∪B∪C)=58
n(F∩B∩C)=3
We know, n(F∪B∪C)=n(F)+n(B)+n(C)−n(F∩B)−n(B∩C)−n(F∩C)+n(F∩B∩C)
n(F∩B)+n(B∩C)+n(C∩F)=73−58+3=18
the number of students who received medals in exactly two of the three
sports is,
n(F∩B)+n(B∩C)+n(C∩F)−3×n(F∩B∩C)=18−9=9
Solution :
Let,
V = set of students getting medals in volleyball
F = set of students getting medals in football
B = set of students getting medals in basket ball
We are given the following:
n ( V ) = 38,
n ( F ) = 15
n ( B ) = 20
n ( V U F U B ) = 58
n ( V ∩ F ∩ B ) = 3
Let,
a = Number of students getting medals in volleyball and football but not the basketball
b = Number of students getting medals in football and basketball but not volleyball
c = Number of students getting medals in basketball and volleyball but not football
So,
n ( V U F U B ) is as follows :
⇒ [ 38 - ( a + c + 3 ) ] + [ 15 - ( a + b + 3 ) ] + [ 20 - ( b + c + 3 ) ] + a + b + c = 58
⇒ 67 - ( a + b + c ) = 58
⇒ a + b + c = 9
∴ Number of players who received medals in exactly two of the three sports,
⇒ a + b + c
⇒ 9
∴ Number of players who received medals in exactly two of the three sports is 9 respectively.
