20 students in class.
There are 10 benches with each bench able to sit two students.
A test were given and student scored 12 A's and 8 B's
8 A's from same bench and 4 A's from students sitting at different bench.
4 B's from the same bench and 4 B's from students sitting at different bench.

What is the probability of this happening (8 A's and 4 B's from the same bench given 12 A's and 8 B's)?

  • $\begingroup$ You might be able to adapt the solution to a very similar question: math.stackexchange.com/questions/236802/… $\endgroup$ – Ross Millikan Nov 14 '12 at 4:06
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    $\begingroup$ Subjective assessments such as "easy", "difficult", "tricky" vary widely -- in my view they contribute very little to summarizing the question for someone scanning the titles. Also, note that the tags are visible wherever the title is visible, so having the only tag as the only substantial word in the title makes it highly redundant. Please consider replacing the title by something more specific, such as "probability of arrangement of pairs of students on benches". $\endgroup$ – joriki Nov 14 '12 at 4:22

Presumably we are to assume that "getting an A" or "getting a B" is a fact of nature that can only be changed by copying. The real situation is more complex.

To make the calculation more neutral, assume that by A people we mean the $12$ people whose names begin with A, and by B people we mean the $8$ people whose names begin with B. We calculate the probability of the configuration described in the problem, on the assumption of randomness.

Imagine that the seats are numbered $1$ to $20$, with $1$ and $2$ on the same bench, with $1$ on the left, and $2$ on the right. Assume that the remaining seats are numbered similarly.

The people can be assigned seats in $20!$ equally likely ways.

We now count the "favourables," the seat assignments that have $4$ benches with two A people each, and $2$ benches with two B people each, and therefore $4$ mixed benches.

The benches to hold only A people can be chosen in $\dbinom{10}{4}$ ways. For each such choice, the A people to fill them can be chosen in $\dbinom{12}{8}$ ways, and then these can be permuted in $8!$ ways, for a total so far of $\dbinom{10}{4}\dfrac{12!}{4!}$ ways.

For each way of filling the double A benches, the double B benches can be chosen in $\dbinom{6}{2}$ ways, and the people to fill them in $\dbinom{8}{4}$ ways. The people can be permuted in $4!$ ways, for a total of $\dbinom{6}{2}\dfrac{8!}{4!}$.

Finally, fill the $4$ "mixed" benches. For any $4$ given benches, the A and B person combinations to fill them can be chosen in $(4!)^2$ ways. The order in which the chosen people sit at their assigned bench can be changed in $2^4$ ways (for each bench, the A person can be on the left or on the right).

We end up with a total of $$\left[\binom{10}{4}\frac{12!}{4!}\right]\left[\binom{6}{2}\frac{8!}{4!} \right]\left[(4!)^2 2^4 \right].\tag{$1$}$$

Remark: To find the probability, we divide the number in $(1)$ by $20!$. There is a good deal of pleasant cancellation.

There is no reason to think that this number will tell us anything interesting about the likelihood of cheating having occurred. Even if the particular configuration happens to be somewhat unlikely, note that if we toss a fair die $10$ times, the sequence of results we get is very unlikely.


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