Question

1. In a class of 60 students, a survey was conducted, 30 students had applied for Addis Ababa University, 25 students applied for Bahir Dar University and 24 students applied for Wachemo University. 11 students applied for both Addis Ababa and Bahir Dar Universities, 6 applied for both Addis Ababa and Wachemo Universities, 9 applied for both Wachemo and Bahir Dar Universities while 4 applied neither of the aforementioned universities. Find 1.. number of students that applied for all the universities. 2 number of students that applied for at least two of the universities. 3 number of students that applied at most two universities. 4 number of students that applied for Addis Ababa but not Bahir Dar University? Please I need your help?​

Answer 1

1. Consult the linked question.
`n(A\cap B\cap W) = \boxed{3}`.

2. We have

`n(A\cap B) = n(A\cap B \cap W) + n(A\cap B \cap W') \\\\ \implies 11 = 3 + n(A\cap B \cap W') \\\\ \implies n(A\cap B\cap W') = 8`

Similarly we can find

`n(A\cap B' \cap W) = 3`

`n(A'\cap B\cap W) = 6`

`n(A\cap B'\cap W') = 16`

`n(A'\cap B\cap W') = 8`

`n(A'\cap B'\cap W) = 12`

Then the total number of students that applied to at least two of the universities is

`\underbrace{n(A\cap B\cap W') + n(A\cap B'\cap W) + n(A'\cap B\cap W)}_{\text{only 2}} + \underbrace{n(A\cap B\cap W)}_{\text{all 3}} = \boxed{20}`

3. There's a small ambiguity here. Are we interested in students that applied to zero universities? If so, there are

`\underbrace{n(U\setminus(A\cup B\cup W))}_{\text{none}} + \underbrace{n(A\cap B'\cap W') + n(A'\cap B\cap W') + n(A'\cap B'\cap W)}_{\text{only 1}} = \boxed{40}`

If we want students that applied to at least 1 school, we omit the first term and get a total of 36.

4. Split this set of students into those that also applied to Wachemo and those that did not.

`n(A \cap B') = n(A\cap B' \cap W) + n(A\cap B'\cap W') = \boxed{19}`