A family is composed of 2 parents, 3 boys and 4 girls. The family is divided into 3 random trios. Let X be the number of trios that are composed of a parent,a boy and a girl. Calculate E(X). Answer by book: 6:7.
I assumed the probability for every trio is equal. The total amount of trios are 9 choose 3 = 84 trios, and the requested trios have 2*4*3=24 possibilities. Now I am not sure how to continue. Does the order of selecting has importance? Is it like taking balls from a sack?
thanks in advance
By constructing a trio the probability that it will be a trio composed of a parent, a boy and a girl is:$$\frac{\binom{2}{1}\binom{3}{1}\binom{4}{1}}{\binom{9}{3}}=\frac{24}{84}=\frac{2}{7}$$ This matches your calculations.
Compose the trios one-by-one and for $i=1,2,3$ let $X_{i}$ take value $1$ if trio $i$ is composed of a parent, a boy and a girl.
Otherwise let $X_{i}$ take value $0$.
The $X_i$ are not independent, but do have the same probability distribution.
There is no reason to think that e.g. the trio composed as second has more or less probability than the first trio to become a trio with a parent, a boy and a girl.
Then: $$X:=X_{1}+X_{2}+X_{3}$$ is the number of trios that are composed of a parent, a boy and a girl.
With linearity of expectation and symmetry we find:
$$\mathbb{E}X=\mathbb{E}X_{1}+\mathbb{E}X_{2}+\mathbb{E}X_{3}=3\mathbb{E}X_{1}=3P\left(X_{1}=1\right)=3\cdot\frac{2}{7}=\frac{6}{7}$$