Indicator linearity

We have 15 boys and 10 girls, totaling 25 students. The students line up uniformly at random. We want the expected number of adjacent positions where one is a boy and the other is a girl (order BG or GB).

Let $X$ be the number of adjacent boy-girl pairs. For each adjacent pair of positions $(i, i+1)$ with $i = 1,\dots,24$, define an indicator $I_i$ that the two students at those positions are of opposite sexes. Then $X = \sum_{i=1}^{24} I_i$, and by linearity of expectation,

By symmetry, the probability is the same for every adjacent pair; denote it by $p$. Hence $\mathbb{E}[X] = 24 \cdot p$.

Now compute $p$. There are $25 \times 24$ equally likely ordered assignments of two distinct students to the two positions. Favorable cases are (boy, girl) and (girl, boy). There are $15 \times 10 = 150$ ways to choose a boy then a girl, and $10 \times 15 = 150$ ways to choose a girl then a boy, for a total of $300$ favorable ordered pairs. Thus

Alternatively, $\Pr(\text{boy then girl}) = \frac{15}{25} \cdot \frac{10}{24} = \frac{1}{4}$, and $\Pr(\text{girl then boy}) = \frac{10}{25} \cdot \frac{15}{24} = \frac{1}{4}$, so $p = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.

Therefore,