"In an examination 70 % of the candidates passed in History and 50% in Geography and 20% students failed in both the subjects. If 500 students passed in both the subjects, the how many candidates appeared for the exam?"


I am unable to understand how to frame the parameters that I am supposed to take into consideration in a Venn Diagram. A slight guidance about the same about the thinking process will be most certainly welcome.


1 Answer 1


Let us define P as the set that contains all the candidates that appeared for the exam. Then, if we define sets H as the candidates that passed history, G the candidates that passed Geography then we know that $|H|=N*.70$, $|G|=N*.50$ and $|H^{c}\cap G^{c}|=.20*N$. Where we define || as the cardinality of a set and $|P|=N.$ That is N is the total number of candidates. We are given:

$$|H\cap G|=500.$$

Les us Consider how we can break down $P$

$$|P|=|H\cap G| + |H\cap G^c| + |H^c \cap G| + |H^{c}\cap G^{c}|.$$

That is the total number of applicants can be divided into the disjoint sets of:
Passed Both Tests, Failed Both Tests, Passed History Failed Geography, Failed History Passed Geography.

We know: $$|H|=|H\cap G|+|H\cap G^c|=.70* N$$ $$|G|=|H\cap G| + |H^{c}\cap G|=.50*N$$ Since $|H\cap G|=500$, this implies $$|H\cap G^{c}|=.70*N-500$$ $$|H^{c}\cap G|=.50*N-500$$.

If we combine everything we get

$$N=500 +.70*N-500 + .50*N-500 + .20*N$$

Solving for N


  • $\begingroup$ I have one doubt in the given concept. I am actually not able to understand how in certain cases we take intersection of two sets as a product of two sets. I guess in probability or in conditional probability this is a very trivial step to be done. What bars us from taking such a step here. A slightly detailed intuition into the same will be most certainly welcome. $\endgroup$ Oct 17, 2021 at 4:57
  • $\begingroup$ Be careful to differentiate between the probability of an event and the event itself. If the Events $H$ an $G$ where hypothetically independent then $P(H\cap G)=P(H)*P(G)$ that is we can take the product of the individual probabilities to get the joint probability. $H\cap G=H*G$ doesn't make sense unless we are considering potentially the cartesian product. If we are considering cardinality $|H\cap G|= |H||G|$ is not guaranteed. $H\{1,2,3\}$ and $G=\{2,3,4\}$ then $|H|=|G|=3$ but $|H\cap G|=2$. $\endgroup$
    – nvm
    Oct 17, 2021 at 5:08

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