Question 1192208
.
Of the 300 grade 10 students of a certain high school, 
120 joined Mathematics club, 115 joined Science club and 100 joined English club. 
Furthermore, 55 joined Mathematics and Science club, 50 joined Mathematics and English club 
and 40 joined Science and English club. 
Finally, 80 students did not join any of these clubs. 
If a student is selected from this group, find the probability that the chosen student joined

a) all the three clubs.

b) Mathematics or the English club.

c) any of the three clubs.

d) Science club only.

e) Mathematics and Science club but not English club.
~~~~~~~~~~~~~~~~~


<pre>
We have a universal set of 300 students and three its subsets

    M (Math club of 120 students),

    S (Science club of 115 students)

    E (English club of 100 students).


We also have their in-pair intersections

    MS (Math and Science of 55 students),

    ME (Math and English of 50 students),

    SE (Science and English of 40 students).


We also have their triple intersection MSE (Math, Science and English club)


Let x be the number of students in the triple intersection MSE.


Apply the inclusion-exclusion principle.  It says that the union of subsets  (M U S U E)  has

    M + S + E - MS - ME - SE + MSE = 120 + 115 + 100 - 55 - 50 - 40 + x = 190 + x  students.    (1)


Taking into account those 80 students who did not joint any of the three clubs, we can write an equation for total students

    300 = 80 + 190 + x,


which gives  x = 300 - 80 - 190 = 30.



So, the triple intersection MSE contains 30 students.


Therefore,  the probability for question (a)  is  {{{30/300}}} = {{{1/10}}} = 0.1.



For question (b), the number of the students in the union  (M U E)  is  M + E - ME = 120 + 100 - 50 = 220 - 50 = 170

Therefore, the probability for question (b) is  {{{170/300}}} = {{{17/30}}}.



For question (c), the number of students in the union  (M U S U E)  is  190 + x = 190 + 30 = 220 students  (see the formula (1)  above).

Therefore, the probability for question (c) is  {{{220/300}}} = {{{11/15}}}.



For question (d), the "Science only" community has  (S \ MS \ SE + MSE) = (115 - 55 - 40 + 30) = 50 students.

Therefore, the probability for question (d) is  {{{50/300}}} = {{{1/6}}}.



For question (e), the "Mathematics and Science club but not English club" community is the set (MS \ MSE) and has  MS - MSE = 55 - 30 = 25 students.

Therefore, the probability for question (e) is  {{{25/300}}} = {{{1/12}}}.
</pre>

Solved.


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On inclusion-exclusion principle, &nbsp;see this Wikipedia article


https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle



To see many other similar &nbsp;(and different) &nbsp;solved problems, &nbsp;see the lessons


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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Advanced-probs-counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Advanced problems on counting elements in sub-sets of a given finite set</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Challenging-problems-on-counting-elements-in-subsets-of-a-given-finite-set.lesson>Challenging problems on counting elements in subsets of a given finite set</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Selected-problems-on-counting-elements-in-subsets-of-a-given-finite-set.lesson>Selected problems on counting elements in subsets of a given finite set</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Inclusion-Exclusion-principle.lesson>Inclusion-Exclusion principle problems</A> 


in this site.



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