Question 1205816: Among 70 Jambites, a survey shows that: 23 were taking Physics 25 were taking Biology 22 were taking Chemistry 6 were taking Physics and Biology 7 were taking Biology and Chemistry 8 were taking Chemistry and Physics 2 were taking Physics, Biology and chemistry. (a) How many of the Jambites were taking none of the three sciences? (b) How many were taking just one of the three subjects?
Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website! .
Among 70 students, a survey shows that:
23 were taking Physics, 25 were taking Biology, 22 were taking Chemistry
6 were taking Physics and Biology, 7 were taking Biology and Chemistry, 8 were taking Chemistry and Physics
2 were taking Physics, Biology and chemistry.
(a) How many of the students were taking none of the three sciences?
(b) How many were taking just one of the three subjects?
~~~~~~~~~~~~~~~~~~~~~~~
You are given the universal set U of 70 students and 3 its basic subsets P, B and C
(see the table below).
U 70 total students
P 23 Physics
B 25 Biology
C 22 Chemistry
Also, you are given info about their in-pair intersections and about their triple intersection.
PB 6 Physics and Biology
BC 7 Biology and Chemistry
PC 8 Chemistry and Physics
PBC 2 Physics, Biology and chemistry
Having this info well organized, you can easily answer all questions (a), and (b).
(a) The set of students not taking any of the three subjects is U \ (P U B U C).
So, calculate the number of students in the union (P U B U C} first.
For it, use the inclusion-exclusion princuple/(formula)
n(P U B U C) = n(P) + n(B) + n(C) - n(PB) - n(BC) - n(PC) + n(PBC) =
= 23 + 25 + 22 - 6 - 7 - 8 + 2 = 51.
Now the next step gives the answer to question (a) :
the number of students taking none of the three sciences = 70 - 51 = 19. ANSWER
(b) The number of students taking Physics only, n(Po), is
n(Po) = n(P) - n(PB) - n(PC) + n(PBC) = 23 - 6 - 8 + 2 = 11.
The number of students taking Biology only, n(Bo), is
n(Bo) = n(B) - n(PB) - n(BC) + n(PBC) = 25 - 6 - 7 + 2 = 14.
The number of students taking Chemistry only, n(Co), is
n(Co) = n(C) - n(PC) - n(BC) + n(PBC) = 22 - 8 - 7 + 2 = 9.
The number of students taking just one of the three subjects is
n(Po) + n(Bo) + n(Co) = 11 + 14 + 9 = 34.
Solved.
----------------
On inclusion-exclusion principle, see this Wikipedia article
https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
To see many other similar (and different) solved problems, see the lessons
- Counting elements in sub-sets of a given finite set
- Advanced problems on counting elements in sub-sets of a given finite set
- Challenging problems on counting elements in subsets of a given finite set
- Selected problems on counting elements in subsets of a given finite set
- Inclusion-Exclusion principle problems
in this site.
Happy learning (!)
|
|
|