The "10" of "grade 10" will be completely ignored because "grade 10" is a categorical label and not a count of any number of students.
We could easily replace the label "grade 10" with "sophomore" to help avoid such a distraction.
Define these sets
U = universal set (collection of all 300 sophomores)
M = set of people in the math club
S = set of people in the science club
E = set of people in the english club
From that we can say
n(U) = number of sophomores total
n(M) = number of people in the math club
n(S) = number of people in the science club
n(E) = number of people in the english club
Let's break down what your teacher gave you into separate bite-sized facts.
- Fact 1 : 300 sophomore students total
- Fact 2 : 120 in the math club
- Fact 3 : 115 in the science club
- Fact 4 : 100 in the english club
- Fact 5 : 55 in math and science
- Fact 6 : 50 in math and english
- Fact 7 : 40 in science and english
- Fact 8 : 80 in neither of the three clubs
| Statement | Reason |
| n(U) = 300 | Fact 1 |
| n(M) = 120 | Fact 2 |
| n(S) = 115 | Fact 3 |
| n(E) = 100 | Fact 4 |
| n(M and S) = 55 | Fact 5 |
| n(M and E) = 50 | Fact 6 |
| n(S and E) = 40 | Fact 7 |
| n(not M, not S, not E) = 80 | Fact 8 |
Now use the inclusion-exclusion principle to find how many people are in all three clubs simultaneously.
n(U) = n(M) + n(S) + n(E) - n(M and S) - n(S and E) - n(M and E) + n(M and S and E) + n(not M, not S, not E)
300 = 120 + 115 + 100 - 55 - 40 - 50 + n(M and S and E) + 80
300 = 270 + n(M and S and E)
n(M and S and E) = 300-270
n(M and S and E) = 30
There are 30 sophomores in all three clubs at once.
Then we can say
n(M and S, not E) = n(M and S) - n(M and S and E)
n(M and S, not E) = 55 - 30
n(M and S, not E) = 25
For more info, check out the concepts about set subtraction and set complements.
Similarly,
n(M and E, not S) = n(M and E) - n(M and S and E)
n(M and E, not S) = 50 - 30
n(M and E, not S) = 20
And,
n(S and E, not M) = n(S and E) - n(M and S and E)
n(S and E, not M) = 40 - 30
n(S and E, not M) = 10
Once again, we'll use the inclusion-exclusion principle to get the following:
n(M only) = n(M) - n(M and S) - n(M and E) + n(M and S and E)
n(M only) = 120 - 55 - 50 + 30
n(M only) = 45
Furthermore,
n(S only) = n(S) - n(M and S) - n(S and E) + n(M and S and E)
n(S only) = 115 - 55 - 40 + 30
n(S only) = 50
Lastly,
n(E only) = n(E) - n(M and E) - n(S and E) + n(M and S and E)
n(E only) = 100 - 50 - 40 + 30
n(E only) = 40
Summary:
- n(M only) = 45
- n(S only) = 50
- n(E only) = 40
- n(M and S, not E) = 25
- n(M and E, not S) = 20
- n(S and E, not M) = 10
- n(M and S and E) = 30
- n(not M, not S, not E) = 80
Take careful note how the numbers are placed in the eight separate regions.
For example, the "30" in the very middle is in all three circles at once to show the number of students who signed up for all three clubs.
Then something like "25" is only in the math and science circles, but outside the english circle.
The other values are handled the same way.
As a way to check...
- Adding up all the numbers in the M circle should get you 120.
- Adding up all the numbers in the S circle should get you 115.
- Adding everything in the E circle should yield 100.
- Adding all of the values should get you 300.
https://www.algebra.com/algebra/homework/Subset/Subset.faq.question.1192015.html
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There's quite a bit of set up required to get the venn diagram.
However, once we have the diagram, the questions can be quickly answered.
For part (a), we want to know the probability of someone being in all three clubs.
The venn diagram shows 30 such people and this is out of 300 total.
30/300 = 1/10 is the probability of getting someone who is in all three clubs at once.
Now onto part (b). Add up the numbers that are in circle M or circle E, or both regions.
45+25+20+30+40+10 = 170
This is out of 300 total, so we get 170/300 = 17/30 as the probability.
For part (c), we add up all the numbers in any of the three circles.
45+25+50+20+30+10+40 = 220
Or as a shortcut, we can note that if 80 people are not in any of the three clubs, then 300-80 = 220 are in at least one club.
220/300 = 11/15
For part (d), we found that there are 50 students signed up for the science club only.
50/300 = 1/6
In the final part (e), we once again refer to the venn diagram.
Locate the value in circles M and S, but outside circle E. This is the value 25.
25/300 = 1/12
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Answers:
a) 1/10
b) 17/30
c) 11/15
d) 1/6
e) 1/12