Probability of selecting two students such that they are in same class Question is:
A group of 60 students is randomly split into 3 classes of equal size. All partitions are equally likely. Jack and Jill are two students belonging to that group. What is the probability that Jack and Jill will end up in the same class.
Correct Answer: 19/59
I was thinking it in following way. Please correct me.
Total no of ways in which two students can be selected : 60C2
ways of choosing jack : 60
ways of choosing Jill : 19 as Jack and Jill mist be in same class
ways of choosing two students such that they are in same class: 60*19
So Answer with this is : 60*19/60C2 = 60*19*2/(60*59) = 19*2/59
 A: Here is another solution using counting. There are ${60 \choose 20}{40 \choose 20}$ ways of partitioning students into class 1,2,3. Think Jill and Jack as one person. There will be $3{58 \choose 18}{40 \choose 20}$ ways of grouping them (e.g. put J-J in class 1, choose $18$ more people, then partition the remaining into $20$-$20$, similarly do this three times for each class) The ratio is $19/59$.
In your solution, you need to consider the denominator as $60\times 59$ since choosing positions for Jack & Jill is without symmetry if you use $C(60,2)$. But, the numerator counts with symmetry, therefore doubled.
A: Here is a simpler Solution. Basically you have 60 seats in total and 3 classes assume A,B,C with each 20 seats. Jack can get 1 seat in any class with Probability 20/60 = 1/3. Suppose that class is A.
Now for Jill to be in class A, there are 19 choices left out of 59. So, probab of Jill sitting in class A (given Jack is in class A) is 19/59.
So, Prob[both belonging to the same specific class(A or B or C)] = 1/3 *(19/59)
Hence, Prob[ both belong to the same class] = 3* (1/3) *(19/59) = 19/59 (as there are total 3 classes)
