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.
= 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

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

$P(\text{"X drinks tea"}) < P(\text{"X drinks tea"} \cap \text{"X lives in hostel"})$

which would violate the properties of probability.

  • $\begingroup$ @Shayan I think that, if this is an assignment, your best bet is to present this as the answer but include a solution where the 60% is the probability for drinking tea and 50% is the probability for both. It might be the original intent of the author. $\endgroup$ – kviiri Nov 17 '16 at 12:19
  • $\begingroup$ If I change the probabilities as you are saying then what is the answer and method of solving it? $\endgroup$ – Shayan Nov 17 '16 at 12:29
  • $\begingroup$ @Shayan The method you're using in your question looks correct, only the initial values are off. $\endgroup$ – kviiri Nov 17 '16 at 12:35

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