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 have enrolled in exactly one test series.?

Option 4 : 30

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 (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.

No. of students who have enrolled only in Navy test series is given by say X.

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

⇒ X = 25 – 9 – 11 + 3 = 8.

No. of students who have enrolled only in NDA test series is given by say Y.

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

⇒ Y = 26 – 11 – 8 + 3 = 10

Similarly, No. of students who have enrolled only in NDA test series is given by say Z.

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

⇒ Z = 26 – 9 – 8 + 3 = 12

So, the number of students who have enrolled in exactly one test series = X + Y + Z = 8 + 10 + 12 = 30.