3
$\begingroup$

Textbook problem: A teacher with a math class of 20 students randomly pairs the students to take a test. What is the probability that Camilla and Cameron, two students in the class, are paired with each other?

My answer: 1/190. There is only one such pair out of the 20-choose-2 possible pairings.

Their answer: 1/19. Camilla can be paired with 19 students and only one such pairing is with Cameron.

Question: What principle am I missing in my reasoning that would help me to see why their answer is right and mine wrong? It's like I follow their line of reasoning too but don't see what underlying assumption differentiates the answers to see how I can frame the problem correctly on my own.

Textbook: Chapter 26 from The Art of Problem Solving (Volume I) by Rusczyk and Lehoczky.

$\endgroup$
1
  • 1
    $\begingroup$ But there are 10 pairs of students. That means Cameron and Camilla get 10 chances for their pair to be chosen. So the probability is definitely higher than 1/190. $\endgroup$
    – eyeballfrog
    May 25 at 20:46

3 Answers 3

Reset to default
4
$\begingroup$

You're missing that the teacher will pair up all the students, making 10 pairs. Camilla and Cameron could be any of these 10.

$\endgroup$
3
  • 2
    $\begingroup$ I want to make the argument 10 pairs / 190 possible pairs = 1/19, but this doesn't work because not all choices of 10 pairs have no repeat students. Is there a variation of this argument that does work? $\endgroup$
    – eyeballfrog
    May 25 at 20:55
  • 5
    $\begingroup$ @eyeballfrog Yes: suppose the pairs are numbered (perhaps by the desk they are sent to). Then the probability Camilla and Cameron are both picked for desk $1$ is $\frac1{190}$ as you found. Similarly desk 2, or each of the other desks. Then use linearity of expectation $\endgroup$
    – Henry
    May 25 at 21:16
  • $\begingroup$ There are 190 total possible pairs, so 190-choose-10 total possible testing classrooms that could be set up by the teacher. Out of all those testing classrooms there are 189-choose-9 testing classrooms you could choose that contain the Camilla-Cameron pair; hence the probability is 189-choose-9 / 190-choose-10 which equals 1/19. $\endgroup$
    – Ungar Linski
    May 26 at 1:19
3
$\begingroup$

The 190 comes from counting all the possible pairs of students, including those that are not Cameron or Camilla. You need to zero in on only Cameron or Camilla pairs. Camilla will be paired with someone, and that is a certainty. She cannot be paired with herself, so that leaves only 19 other students she can be paired with.

If instead of pairing up everyone, you only picked two students at random to make one pair out of the 20, then the odds would be 1/190.

$\endgroup$
0
$\begingroup$

$\mathbf{\text{For another approach:}}$

Let assume that the boxes represent groups such that two adjacent boxes are in the same group. $$\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad$$

Now , let Camilla select one of the position , she can do it $\frac{20}{20}=1$ ways.After that Cameron must select an empty space , but we want that they are in the same group , so if we assume that Camilla selected the position $\color{red}{1}$ , then Cameran must select position $2$ among $19$ possible empty position. As you see , the probability that Cameron select the position $2$ is $1/19$ , then the the probability that they are in same group is $1/19$ $$\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{$\color{red}{1}$}\fbox{2} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad\fbox{+}\fbox{+} \quad$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.