I have a following question from a GRE test prep book:

In a probability experiment, $G$ and $H$ are independent events. The probability that $G$ will occur is $r$, and the probability that $H$ will occur is $s$, where both $r$ and $s$ are greater than $0$.

In the answer, the probability that either $G$ will occur or $H$ will occur, but not both is defined as: $r + s - 2rs$.

I thought the union of two events A and B is given as $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ which in this question equates to $r+s-rs$. My questions are: Which one is correct? Does the definition of union include the probability that both events will occur?

  • $\begingroup$ Well note that for independent events you have $P(A)P(B)=P(A)\cap{}P(B)$ so both are correct. The probability that both will occur is implied by the inclusion exclusion principle. $\endgroup$ – Μάρκος Καραμέρης May 30 '18 at 23:25
  • $\begingroup$ Do not confuse the phrase "A or B (which includes possibility of both)" with the phrase "Either A or B (but not both)". The first event is represented by $A\cup B$. The second event is represented by $A\triangle B$ or as $(A\setminus B)\cup (B\setminus A)$ or as $(A\cup B)\setminus(A\cap B)$. The question asks you to find the probability of this second phrase, not the first. $\endgroup$ – JMoravitz May 30 '18 at 23:30
  • $\begingroup$ If you want to be pedantic, the phrase "Either A or B" should strictly imply not both, but sadly language evolves and people misuse phrases so it unfortunately is used by people incorrectly to mean "A or B." This problem however strictly points out the not both aspect. Computer scientists might use "XOR" in place of "Either...or" but it is not common yet to hear it in conversation. $\endgroup$ – JMoravitz May 30 '18 at 23:33
  • $\begingroup$ @JMoravitz Thank you! Out of curiosity, what is the mathematical term for $A△B$? $\endgroup$ – jellomutiny May 30 '18 at 23:36
  • $\begingroup$ The symmetric difference of $A$ and $B$. $\endgroup$ – JMoravitz May 30 '18 at 23:37

Yes, the definition of union includes cases in which both occur. Thus both statements are correct: the probability of $A \cup B$ (by definition including cases both occur) is $r+s-rs$, and to get the probability that one but not both occurs, subtract the probability that both occur ($rs$) to get $r+s-2rs$

  • $\begingroup$ Makes sense thank you! $\endgroup$ – jellomutiny May 30 '18 at 23:41

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