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
(a) Rebecca and Elise will be paired?
(b) Rebecca and Elise will be chosen to represent their schools but will not play each other?
(c) either Rebecca or Elise will be chosen to represent her school?
I have seen some online resources and tried solving this question, but have been unable to.
For part $(a)$, I get the required numerator by fixing Rebecca and Elise, then permuting the rest by $\binom{7}{3} \cdot \binom{8}{3} \cdot 3!$ and the denominator for total possible pairings as $\binom{8}{4} \cdot \binom{9}{4} \cdot 4!$. This returns $1/18$.
I don't know how to solve part (b) and (c). I tried the (Total ways - number of ways where both sisters are together) using: $\binom{8}{4}\cdot\binom{9}{4}\cdot4! - \binom{7}{3}\cdot\binom{8}{3}\cdot3!$ divided by the total $\binom{8}{4}\cdot\binom{9}{4}\cdot4!$ but is so different from the book.
I am also not sure if the solutions are wrong or the text book, or me. I only know permutation & combinations, and not aware of random variables, conditional probabilities, or bayes theorem, or other ways. I am not even sure if selecting with replacement (the one used in picking colored bals from urns) will be used here or not. Very confused and helpless. So kindly help me with the way how I should approach this question. I will really appreciate it.
(a) The probability that Rebecca will be chosen is $P_1=\frac{4}{8}=\frac{1}{2}$.
The probability that Elise will be chosen is $P_2=\frac{4}{9}$.
The probability that they both will be chosen is $P_1\times P_2=\frac{2}{9}$.
In case they both are chosen, they will have $\frac{1}{4}$ chance to be paired. So, the answer of (a) is $\frac{2}{9}\times \frac{1}{4}=\frac{1}{18}$.
(b) In case they both are chosen, they will have $\frac{3}{4}$ chance NOT be paired. So, the answer of (b) is $\frac{2}{9}\times \frac{3}{4}=\frac{1}{6}$.
(c) The probability that either Rebecca or Elise will be chosen is
$1-(1-P_1)\times(1-P_2)=1-\frac{1}{2}\times \frac{5}{9}=\frac{13}{18}$.