Probability of two friends sit one near the other

Question

Probability of two friends sitting together

Hi,

I have the following problem and I would like to ensure that I solved it correctly. In a class-room there are 4 rows of chairs, 3 chairs per raw. The teacher randomly allocates 12 pupils to 12 possible chairs. John and Jack are two friends. What is the probability that they will sit one near the other (i.e., the adjacent chairs of the same row)?

My solution: 1) For each row, there are 2 solutions that satisfy John and Jack (unordered; I do not care where each one sits). There are 4 rows. So, number of satisfactory solutions is: 2 x 4 = 8

2) Total number of possibilities John and Jack might sit (unordered sampling) is: 12! / 10! * 2 ! = 66.

3) Probability is: 8 / 66 = 0.121

Am I right?

Many thanks.

Answer 1

Your result is right. My thoughts were the following. The probability that the two friends are sitting next to each other at chair i and j is $p_{ij}=\frac1{11}\cdot \frac1{12}$

Now we count the number of favorable arrangements i and j in a row.

$(1,2);(2,1);(2,3);(3,2)\Rightarrow \texttt{4 arrangements in a row}$

Thus the probability that John and Jack are sitting next to each other is

$$\frac{4\cdot 4}{11\cdot 12}=\frac8{66}=0.\overline{12}\approx 12\%$$

Answer 2

You are correct.

Two-thirds of the $12$ students are at row-ends and have exactly one neighbor. For those students, there is a $\frac1{11}$ chance their special friend is that neighbor. The other third of the students (in center seats) have two neighbors, and there is a $\frac2{11}$ chance their special friend is one of those neighbors. Thus the total probability a student has their special friend as a neighbor is $\frac23\cdot\frac1{11}+\frac13\cdot\frac2{11}=\frac4{33}$