The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that either Rebecca or Elise will be chosen to represent her school?

Here is what I tried to do; Event A is probability that Rebbecca gets chosen to represent her school, which is $\frac{4}{8}$ and the same goes for event B, which probability that Elisa is chosen to represent her school is $\frac{4}{9}$. Now for the interception where both events occur at the same time is $\frac{4}{8}$*$\frac{4}{9}$.

$P(A\cup B)$ = $P(A) + P(B) - P(AB)$ = $\frac{4}{8}$ + $\frac{4}{9}$ - ($\frac{4}{8}$*$\frac{4}{9}$) = $\frac{13}{18}$

However, according to my book this is not the right answer and it should have been $\frac{1}{2}$ instead.

Could anyone provide some insight on this kind of them. And a clear way to solve it. Thanks.


What is probably the case is that when the question asked for the probability of "either Rebecca or Elise", it is intending the exclusive or case, where it is one or the other, but not both. This becomes: $$ P(A \cap \bar{B}) \cup P(\bar{A} \cap B) $$ Which would be: $$ \left(\frac{4}{8}\cdot\frac{5}{9}\right) + \left(\frac{4}{8}\cdot\frac{4}{9}\right) = \frac{20}{72} + \frac{16}{72} = \frac{36}{72} = \frac{1}{2} $$

  • $\begingroup$ Why is the complement of B $\frac{5}{9}$ and complement of A is $\frac{4}{8}$ ? $\endgroup$ – Kururugi Suzaku Sep 9 '13 at 3:31
  • $\begingroup$ the complement of a fraction is determined by $1-$ the fraction if the fraction is less than 1 (proper fraction) $\endgroup$ – VikeStep Sep 9 '13 at 3:34
  • $\begingroup$ Because $A$ has $8$ people in her school and $B$ has $9$. If $P(A) = \frac{4}{8}$ then $P(\bar{A}) = 1 - P(A) = 1-\frac{1}{2}=\frac{1}{2}$. Similarly, if $P(B) = \frac{4}{9}$ then $P(\bar{B}) = 1 - P(B) = 1-\frac{4}{9}=\frac{5}{9}$. $\endgroup$ – Avraham Sep 9 '13 at 3:35
  • $\begingroup$ Wow!! I totally forgot about that. Thanks a lot. :) $\endgroup$ – Kururugi Suzaku Sep 9 '13 at 3:55
  • 1
    $\begingroup$ $\frac{4}{8} = \frac{1}{2}$ ;) $\endgroup$ – Avraham Sep 10 '13 at 1:41

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