# Can anybody help me solve this probability related question?

In class 70% students are staying in hostels, 50% loves tea & 60% falls in both categories. What is the probability that a randomly picked student is neither staying in hostel nor loves tea?

This is how I did this:
Event A are students staying in hostels.
Event B are students who loves tea.
P(A)=70/100=0.7
P(B)=50/100=0.5
P(A∩B)=60/100=0.6
P(AUB)=P(A)+P(B)-P(A∩B)
= 0.7+ 0.5- 0.6= 0.6
Student who are neither hostelite nor loves tea will be: 1-P(AUB)=1-0.6=0.4

• This does not make sense: there are more hostelite students who love tea than there are who just love tea. – user228113 Nov 17 '16 at 11:43
• When you correct it using the "edit" button, include your attempt of solution. – user228113 Nov 17 '16 at 11:45

The question is a trick question or there is an error. For any events $A$ and $B$, $P(A \cap B) = P(A) \times P(B|A)$. Since $P(A)$ and $P(B|A)$ are probabilities, they are in the range $[0, 1]$, so it is impossible for $P(A \cap B)$ to be greater than $P(A)$ or $P(B)$. However, the question presents that for a randomly selected student $X$,
$P(\text{"X drinks tea"}) < P(\text{"X drinks tea"} \cap \text{"X lives in hostel"})$