In a group of 50 persons, 14 drink tea but not coffee and 30 drink tea.

Solution:

Let A & B denote the sets of the persons who drink tea & coffee, respectively .

Given:

$n(A \cup B)=50$

$n(A)=30$

 

$n(A-B)=14$

(i) $n(A-B)=n(A)-n(A \cap B)$

$\Rightarrow 14=30-n(A \cap B)$

 

$\Rightarrow n(A \cap B)=16$

Thus, 16 persons drink tea and coffee both.

(ii) $n(A \cup B)=n(A)+n(B)-n(A \cap B)$

$\Rightarrow 50=30+n(B)-16$

 

$\Rightarrow n(B)=36$

We have to find $n(B-A)$

$\Rightarrow n(B-A)=n(B)-n(A \cap B)$

 

$\Rightarrow n(B-A)=36-16=20$

Thus, 20 persons drink coffee but not tea.