Where to sit?

Suppose that the class consists of $n$ boys and $n$ girls and $n$ two-seater desks. For any $1 \le j \le n$ , let $X_j$ be the random variable which is equal to $1$ if the $j^\mathrm{th}$ girl is sitting next to another girl, and equal to $0$ otherwise. Thus the total number of girls who are sitting next to a girl is $Y \; = \; X_1 + X_2 + \cdots + X_n \;.$ Since all seating arrangements are equally likely, any particular student will have any of the other $2n-1$ pupils as a desk partner with equal probability. Thus the probability that the $j^\mathrm{th}$ girl is sitting next to another girl is the probability that she has one of the remaining $n-1$ girls as a desk partner, out of the total number of $2n-1$ possible partners, namely $\frac{n-1}{2n-1}$ . In other words $E[X_j] \; = \; P[X_j = 1] \; = \; \frac{n-1}{2n-1}$ for all $1 \le j \le n$ , and hence the expected number of girls who are sitting next to another girl is $E[Y] \; = \; E[X_1] + E[X_2] + \cdots + E[X_n] \; = \; n \times \frac{n-1}{2n-1} \; = \; \frac{n(n-1)}{2n-1}$ When $n=15$ this expected value is $\frac{210}{29}$ , making the answer $\boxed{239}$ .

My solution is as following: The probability for a girl to sit on the first chair is 1/15, then there are 14 girls left. The probability for another girl to sit on the second chair is now 1/14, so the combined probability is 1/15 + 1/14 = 29/210, then a+b= 239. But my steps do not really obey the info given. I just neglect the information given above about the boys etc, just calculate randomly haha

First we realize that number of desks with only girls must be equal to number of desks with only boys, named k. Then the density of the probability f(k) can be calculated:

$f(k)=\frac{15!^{3}\cdot 2^{15-2*k}}{(15-2*k)!(k)!^{2}\cdot 30!}$

For the expected number of girls we obtain: $\sum_{k=0}^{7}f(k)\cdot 2k= 7.24137931034483$

With happiness we determine that this value can be written as $\frac{210}{29}$ resulting in solution $\boxed{239}$ .

Let $I_{i}$ denotes the indicator variable: it's 1 if the $i^{th}$ table have two girls and 0 otherwise. Based on the question, we want to know $\sum_{i=1}^{15}E[2I_{i}]$ . Since $E[I_{i}] = \frac{C_{15}^{2}}{C_{30}^{2}} = \frac{7}{29}$ , we have $\sum_{i=1}^{15}E[2I_{i}] = 15\times 2 \times \frac{7}{29} = \frac{210}{29}$ . Thus a + b = 210 + 29 = 239