In a survey of 100 students, it was found that 25 students have enrolled for Navy AA test series, 26 students have enrolled for NDA test series, 26 students have enrolled for Airforce group X test series, 9 students have enrolled for both Navy AA and Airforce group X test series, 11 students have enrolled for both Navy AA and NDA test series, 8 students have enrolled for both NDA and Airforce group X test series whereas 3 students have enrolled in all three test series. Find the number of students who has not enrolled in any of the test series.?

Option 3 : 48

Concept:

Let A, B and C be three finite sets and U is the finite universal set, then

1. n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
2. n (A ∪ B) = n (A) + n (B) ⇔ A ∩ B = ϕ
3. n (A - B) = n (A) – n (A ∩ B) = n (A ∩ B’)
4. n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C)
5. n (A’ ∪ B’) = n = n (U) – n (A ∩ B)
6. n (A’ ∩ B’) = n = n (U) – n (A ∪ B)
7. n (A Δ B) = n (A) + n (B) – 2 n (A ∩ B)
8. n (A’) = n (U) – n (A)

Calculation:

Let, A = No. of students enrolled in Navy AA test series, B = No. of students enrolled in NDA and C = No. of students enrolled in Airforce goup X test series.

Given: n (U) = 100, n (A) = 25, n (B) = 26, n (C) = 26, n (A ∩ C) = 9, n (A ∩ B) = 11, n (B ∩ C) = 8 and n (A ∩ B ∩ C) = 3.

As we know that, if A, B and C are 3 finite sets then

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C)

⇒ n (A ∪ B ∪ C) = 25 + 26 + 26 – 11 – 8 – 9 + 3 = 52

So, there are 52 students who enrolled in any of the 3 test series.

The number of students who has not enrolled in any of the test series is given by say X.

X = n = n (U) – n (A ∪ B ∪ C)

⇒ X = 100 – 52 = 48.