The chess clubs of two schools consist of, respectively, $8and9$ 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?

Given in the question that The chess clubs of two schools consist of, respectively, $8and9$ 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 we have to find that The probability that Rebecca and Elise will be paired The probability that Rebecca and Elise will be paired.

The probability that Rebecca will be chosen to represent her school is $\frac{4}{8}$.

The probability that Elise will be chosen to represent her school is $\frac{4}{9}$.

Assuming that both Rebecca and Elise are chosen, The probability that The probability that they will be paired is $\frac{1}{4}$

Therefore, the probability that Rebecca and Elise will be paired is localid="1650280038160" $\frac{4}{8}.\frac{4}{9}.\frac{1}{4}=\frac{1}{18}$

Given in the question that The chess clubs of two schools consist of, respectively, $8and9$ 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 we have to find that The probability that Rebecca and Elise will be chosen to represent their schools but will not play each other.

The probability that Rebecca will be chosen to represent her school is $\frac{4}{8}$

The probability that Elise will be chosen to represent her school is $\frac{4}{9}$.

Assuming that both Rebecca and Elise are chosen, the probability that they will not be paired is $\frac{3}{4}$.

Therefore, the probability that Rebecca and Elise will be chosen to represent their schools but will not play each other is$\frac{4}{8}.\frac{4}{9}.\frac{3}{4}=\frac{1}{6}$

Given in the question that The chess clubs of two schools consist of, respectively, $8and9$ 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 we have to find The probability that either Rebecca or Elise will be chosen to represent her school .

Let $R$be the event that Rebecca is chosen to represent her school and let Elise be the event that $E$is chosen to represent her school.

We are interested in calculating $P(R\cup E)$. By the inclusion-exclusion identity, we have

localid="1650014052113" $$\begin{gathered} P(R\cup E)=P\left(R\right)+P\left(E\right)-P\left(RE\right) \\ =\frac{4}{8}+\frac{4}{9}-\left(\frac{4}{8}.\frac{4}{9}\right) \\ =\frac{13}{18} \end{gathered}$$