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 the 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 Rebecca and Elise will be paired?

The answer is $\displaystyle\frac48 \frac49 \frac 14$. How did they get the $\dfrac14$? Wouldn't there be $16$ possible matches, one of which would be sister vs sister, so shouldn't it be $\displaystyle\frac48 \frac49 \frac1{16}$?

New question (second part): What is the probability that either Rebecca or Elise will be chosen to represent her school?

Originally, this did not confuse me but now it does. I used this formula: P(E) + P(R) - P(E and R) = 4/8 + 4/9 - 4/8*4/9 = 13/16 (first method). It did not match the answer in the book which is 1/2 but it seemed to make sense. Now I've seen another approach: 4/8 * 5/9 + 4/8 * 4/9 = 1/2 (second method), which matches the answer in the book. Which is correct and if the second method is correct, why is the second method correct and the first wrong?

  • $\begingroup$ You don't have all 16 possible matches, only 4 is actually played. And the other school has 9 players. $\endgroup$ – user10354138 May 18 at 16:49
  • $\begingroup$ @o-roi please mark correct answers in stack overflow for two reasons: 1. whom that look at question and answers. 2, for increasing scores of responsive. Thanks $\endgroup$ – BarzanHayati May 19 at 13:13

First off, Rebecca and Elise need to be chosen in order to be able to get matched up. Both of them together are chosen with a probability of $\frac48\cdot\frac49$.

Now that both of them have been chosen into the respective teams, there are only $\mathbf4$ players that Elise could get matched up with (note that every player from one team is matched up with one random player of the second team. They don't randomly choose a single player from each team and call it a day.) Thus the probability of Rebecca getting matched up with Elise is $\frac14$. The answer key is thus correct.

  • $\begingroup$ But the OP has clearly mentioned that The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. How is this the same as every player from the first team is matched up with every player of the second team.? I'm confused, could you please elaborate here? $\endgroup$ – ExtremeRaider May 18 at 17:14
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    $\begingroup$ There are 4 pairings that are dependent on each other. This is exactly what my answer assumes. I changed the ambiguity with using the word every twice: What I meant to say is that each player from one team gets matched up with exactly one random player from the other team. $\endgroup$ – Maximilian Janisch May 18 at 17:15
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    $\begingroup$ Alright, that clears it up. Thanks! $\endgroup$ – ExtremeRaider May 18 at 17:16
  • $\begingroup$ @ExtremeRaider You are welcome :) $\endgroup$ – Maximilian Janisch May 18 at 17:17
  • $\begingroup$ Thanks for your response. It makes it clear that it is 1/4 because a particular player can only be matched with 1 of the other 4 players on the other team. Out of curiousity, what sort of wording would be used so that it would be 1/16 instead of 1/4? It seems like in dealing with probability problems the way things are worded can affect the problem. $\endgroup$ – o Roi May 18 at 19:25

All of possible state for having match between four player from first team and four from second team will be $$\binom{8}{4} \binom{9}{4} 4!$$ ($4!$ for matching between two teams).

Now for selecting Rebeca and Elice for first and second team we must select other 3 people: so we have $$\binom{1}{1}\binom{7}{3} \binom{1}{1}\binom{8}{3}$$ and $3!$ for matching other 3's). So we have $$\binom{1}{1}\binom{7}{3} \binom{1}{1}\binom{8}{3} 3!$$ possible states. be dividing these values, we reach to $$\frac{\binom{1}{1}\binom{7}{3} \binom{1}{1}\binom{8}{3} 3!}{\binom{8}{4} \binom{9}{4} 4!}=\frac{4*4*1}{8*9*4}=\frac{1}{18}$$


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