Explanation
Solution
Let the three sets be M (Mathematics), S (Statistics), and P (Physics).
1. Given Data:
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Total students n(U)=100
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Mathematics only =15
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Statistics only =12
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Physics only =8
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Mathematics and Statistics n(M∩S)=10
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Physics and Mathematics n(P∩M)=40
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Physics and Statistics n(P∩S)=20
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Total Physics n(P)=65
2. Finding the Intersection of all three (M∩S∩P):
Let the number of students who offered all three subjects be x.
From the data of the Physics set, we can break down the total (65) into its components:
n(P)=(Physics only)+(Physics and Math only)+(Physics and Stats only)+(All three)
Substitute these into the Physics total:
65=8+(40−x)+(20−x)+x
65=68−x
x=68−65=3
So, 3 students offered all three subjects.
3. Finding the Total number of students who offered at least one subject:
Now we calculate the other intersection parts:
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Math and Stats only =n(M∩S)−x=10−3=7
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Physics and Math only =40−3=37
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Physics and Stats only =20−3=17
Total students who offered at least one subject n(M∪S∪P):
=(Math only)+(Stats only)+(Physics only)+(Math & Stats only)+(Phys & Math only)+(Phys & Stats only)+(All three)
=15+12+8+7+37+17+3
=99
4. Students not offered any of the three:
Not offered any=Total students−n(M∪S∪P)
Correct Option: (a)
