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I stumbled upon following probability problem from Sheldon Ross' book:

The chess clubs of two schools consists 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) exactly one of Rebecca and Elise will be chosen to represent her school?

I was not able to solve the problem. So I referred to various answers online. Out of them all, I understood solution only for first one. I found two approaches. I understood both approaches for problem (a):

Problem (a) Approach 1

There are 4 options for boards at which sisters will sit. On remaining three boards, a team of remaining three players from club 1 can be selected and made to sit in ${}^7P_3=7\times6\times5$ ways and a team of remaining three players from club 2 can be selected and made to sit in ${}^8P_3=8\times7\times6$ ways. So total number of ways sister get paired together $=4\times{}^7P_3\times{}^8P_3$. Total number of ways in which team of four players from both clubs can be made to sit across boards $={}^9P_4\times{}^8P_4$. Thus the final probability $=\frac{4\times{}^7P_3\times{}^8P_3}{{}^9P_4\times{}^8P_4}=\frac{1}{18}$

Problem (a) Approach 2

After selecting Rebecca and Elise from clubs, we need to select three players from remaining players of each club. This can be done in $\binom{7}{3}\binom{8}{3}$ ways. To determine how these three players from both teams can be paired, we fix some order of three players from one team and permute three players from other team ($3!$ ways). Thus total number of ways in which both sister can get paired together will be $\binom{7}{3}\binom{8}{3}3!$. Similarly total number of ways in which teams can be formed without the restriction that sister should be paired together will be $\binom{8}{4}\binom{9}{4}4!$. Thus the final probability will be $\frac{\binom{7}{3}\binom{8}{3}3!}{\binom{8}{4}\binom{9}{4}4!}=\frac{1}{18}$

But dont follow how similar approach are followed on other problems:

Problem (b) approach 1 (source)

Doubt-1: I didnt get how $12$ comes below:
$=\frac{\color{red}{12}\times{}^7P_3\times{}^8P_3}{{}^9P_4\times{}^8P_4}=\frac{1}{6}$

Problem (b) approach 2 (source)

Doubt-2: I didnt get how $3\times 3!$ comes below:
$\frac{\binom{7}{3}\binom{8}{3}\color{red}{3\times 3!}}{\binom{8}{4}\binom{9}{4}4!}=\frac{1}{6}$

Problem (c) approach 1

  • If $A$ is event that sister from club of $8$ members (club $A$) is chosen and $B$ is event that sister from club of 9 members (club $B$) is chosen, then the desired probability $=P(A\cap B')\cup P(A'\cap B)$.
  • When sister from club $A$ is chosen, remaining can be chosen in ${}^7C_3$ ways. Without restriction of selection of sister, it would have been ${}^8C_4$. Probability that sister from club $A$ is chosen $P(A)=\frac{{}^7C_3}{{}^8C_4}=0.5$.
  • When sister from club $B$ is chosen, remaining can be chosen in ${}^8C_3$ ways. Without restriction of selection of sister, it would have been ${}^9C_4$. Probability that sister from club $B$ is chosen $P(B)=\frac{{}^8C_3}{{}^9C_4}=0.444$.
  • We can find $P(A') = 1-P(A)$ and $P(B') = 1-P(B)$: $(0.5\times (1-0.444))+((1-0.5)\times 0.444))-0.5$

Problem (c) approach 2

Doubt-3: Though I can cross check whether values of fractions given in this answer matches with fractions in above solution, I dont understand how they are obtained directly with any logic. How can we apply logic to obtain these fractions from this solution "directly":

$$\left(\frac{4}{8}\cdot\frac{5}{9}\right) + \left(\frac{4}{8}\cdot\frac{4}{9}\right) = \frac{20}{72} + \frac{16}{72} = \frac{36}{72} = \frac{1}{2}$$

Doubt-4: Solution given on page 2 of this pdf seems totally incorrect right?

Doubt-5: Can I apply any other approaches to get any soluion.

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For Doubt 1, consider that we can first decide where Rebecca sits. We have $4$ boards to choose from, and the remaining boards can be filled by her team members in $^8P_3$ ways. Now to figure out where Elise sits, we know that she can't sit at the same board as Rebecca, so we have $4 - 1 = 3$ options for her board. The remaining $3$ boards are filled by her team members in $^7P_3$ ways. The $12$ comes from the $4 \times 3$ from when we choose Rebecca's board to when we choose Elise's board.

For Doubt 2, we first start by doing $\binom{8}{3}$ and $\binom{7}{3}$ to pick the players on each team that aren't Rebecca or Elise, similar to what we did in part (a). Now, the differences comes in that we don't want Rebecca and Elise to be paired. WLOG, let's look at this from Rebecca's perspective. We want her to play someone that's not Elise, and so we have $3$ choices. Now the condition that Rebecca and Elise won't play is met, so we can fix the matches among the $3$ remaining students from each school using the same logic as in part (a), giving us $3!$. Since Rebecca has $3$ choices of opponent and there will be $3!$ ways to arrange the remaining matches, that's where $3 \times 3!$ comes from.

For Doubt 3, consider that we only want one of Rebecca or Elise to play and not both. There's 2 scenarios in which this can happen, the one in which Rebecca is playing and Elise isn't, and the one in which Rebecca isn't playing and Elise is. To calculate the probability that Rebecca isn't playing and Elise is, we first find the probability that Rebecca is playing. The probability Rebecca is playing is $\frac{4}{9}$ because there are $4$ spots and $9$ members. The probability that Elise isn't playing is $\frac{8 - 4 = 4}{8}$ because there are $8 - 4 = 4$ people without spots out of $8$ total members. We multiply these fractions together because they're independent events. The same idea is followed from the other side where Rebecca isn't selected $\frac{9 - 5 = 4}{9}$ and Elise is $\frac{4}{8}$.

For Doubt 4, they're using the inclusion exclusion principle, which states that $|A \cup B| = |A| + |B| - |A \cap B|$. Let A and B represent Rebecca being selected and Elise being selected, respectively. In the solution you listed, they use the inclusion-exclusion principle correctly, but this doesn't give the intended solution, which is that ONLY ONE of Rebecca or Elise is selected. Instead it gives $|A \cup B|$, which includes the possibility of both Rebecca and Elise being selected. To go from $|A \cup B|$ to $A$ exclusive-or $B$, we need to subtract the probability of $A \cap B$ again, which gives $\frac{13}{18} - \frac{4}{18} = \frac{9}{18} = 0.5$.

For Doubt 5, there are definitely other ways to think about this. Let me display one possible way to think about it that might be more intuitive to you. We again have $2$ cases, the first is that Rebecca is selected and Elise isn't, and the second is that Rebecca isn't selected and Elise is. Let's discuss the first case here and see if you can try to figure out how to solve the second case on your own. For the first case, we need to start by figuring out the probability of Rebecca being selected. There's a total of $\binom{9}{4}$ ways to perform the selections, and there's $\binom{8}{3}$ ways to perform the selections, with Rebecca being selected. The reason for the $8$ and the $3$ is that we give Rebecca a spot, and we need to fill the remaining $4 - 1 = 3$ selections out of the remaining $9 - 1 = 8$ members. For the probability of Elise to not be selected, there's a total of $\binom{8}{4}$ ways to perform the selections, and there's $\binom{7}{4}$ ways to perform the selections with Elise being not selected. The reason for the $7$ and the $4$ is that we want to select from all the students minus Elise, so $8 - 1 = 7$ choices, and unlike in Rebecca's case here, we haven't filled a seat yet, so we still need to select $4$ people. So the probability of case 1: Rebecca selected and Elise selected is: $\frac{\binom{8}{3}}{\binom{9}{4}} \times \frac{\binom{7}{4}}{\binom{8}{4}} = \frac{56}{126} \times \frac{35}{70} = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$. Note that we can simply multiply the probability that Rebecca is selected and that Elise isn't selected here, because Rebecca's selection and Elise's selection are independent events. Now see if you can calculate the other case (Rebecca not selected and Elise is selected) and add them together to get your answer.

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  • $\begingroup$ (Doubt 1) Rebecca can sit on any board. Remaining boards can be filled by Rebecca's team members in ${}^8P_3$ ways. E can sit against any of these team members of R (but not against R). Hence 3 choices. The remaining 3 members of E can be selected and made sit on remaining boards in ${}^7P_3$ ways. So why its not $4\times {}^8P_3\times 3\times {}^7P_3$? I understand this double counts same pairings such as this: (Rabc, lEmn) and (Racb,lEnm), i.e. R paired with l, a with E, b with m and c with n. But then how just $4\times 3$ makes sense? $\endgroup$ – anir Jun 15 '18 at 12:01
  • $\begingroup$ (Doubt 3) I really didnt get "The probability Rebecca is playing is 4/9 because there are 4 spots and 9 members. The probability that Elise isn't playing is (8−4=4)/8 because there are 8−4=4 people without spots out of 8 total members." By "there are 4 spots and 9 members", do you mean "4 chess boards and 9 members"? If yes, then dont know how this takes "Rebecca is playing" into account. In problem (c) approach (1), bullet point 3, you can see I explicitly select 1 sister (say Rebecca), then out of remaining 8, I have to select 3...[continued] $\endgroup$ – anir Jun 15 '18 at 12:02
  • $\begingroup$ [continued]...Making total probability $\frac{{}^8C_3}{{}^9C_4}=0.444=\frac{4}{9}$. But then how "having 4 chess boards and 9 members" equates to same probability as "Rebecca is playing". Same for "4 people without chess boards out of 8" equates to "Elise isnt playing". Is there any simpler logic, that I am missing, which will help me to come up with those probabilities directly without going through those binomials? May be rephrasing may help me understand. $\endgroup$ – anir Jun 15 '18 at 12:03
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    $\begingroup$ I'm not sure what you mean by dropping those terms, can you explain further. $\endgroup$ – Green Jun 15 '18 at 18:25
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    $\begingroup$ Glad that doubt is cleared! For doubt 3, you can maybe think about it this way. Let's consider that we arrange the $9$ students in an ordered line, and we just take the first $4$ students in this ordered line as the students that will play in the chess match. Rebecca is just as likely to be in spot $1$ as she is in spot $2$ or spot $3$ and so on... up to spot $9$, because if we consider all possible ordered arrangements of the students, she will appear an equal number of times in spot $1$ as she does in spot $2$ and so on... Then to appear in the first $4$ spots she has $\frac{4}{9}$ chance. $\endgroup$ – Green Jun 15 '18 at 18:43
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I'm doubtful re the answer(s) Here's my "different" approach which you may find simpler.
Assume Rebecca is in A and Elise in B (it doesn't matter)

a

P(both R and E selected) $\times$ P(R gets E to play with ) $= \frac48\cdot\frac49 \times \frac14 $

b

P(both R and E selected) $\times$ P(R doesn't get E to play with ) $= \frac48\cdot\frac49 \times \frac34$

c

P(only R selected) + P(only E selected) $= \frac48\cdot(1-\frac49) + (1-\frac48)\cdot\frac49$

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  • $\begingroup$ How R gets selected has probability $\frac{4}{8}$ and E gets selected has probability $\frac{5}{9}$. I understand its $\frac{{}^7C_3}{{}^8C_4}=0.5$ (which turns out to be same as $\frac{4}{8}$) and $\frac{{}^8C_3}{{}^9C_4}=0.4444$ (which turns out to be same as $\frac{5}{9}$). But how did you come up with them directly? Any simpler logic, that I am missing, which will help me to come up with those probabilities directly without going through those binomials? $\endgroup$ – anir Jun 15 '18 at 9:58
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    $\begingroup$ If there are 9 numbered marbles and 5 are chosen randomly, any particular marble has $Pr = \frac59$ to be among those chosen. $\endgroup$ – true blue anil Jun 15 '18 at 14:11
  • $\begingroup$ Ok I got it. Thanks. But it seems that you have made mistake. For (a) It should be $\frac{4}{8}.\frac{\color{red}{4}}{9}\times\frac{1}{4}$ which matches with given answer $\frac{1}{8}$, right? (similar for problem (b)) $\endgroup$ – anir Jun 15 '18 at 18:40
  • $\begingroup$ My bad, typos corrected ! btw, the answer is $\frac1{18}$, not $\frac18$ as in your comment. The printer's devil spares no one ! $\endgroup$ – true blue anil Jun 16 '18 at 5:05

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