Binomial law - Another mistake in my workbook ?!?!

Question

A class is divided in $4$ groups, each one made of $4$ boys and $2$ girls.The teacher will choose $1$ representative for each group.

What is the probability that between the $4$ representatives would be exactly $2$ girls?

By the enunciation, it's clear that we are in the presence of binomial law.There is $4$ trials, and in each trial the probability to the event happens is the same.

Let be A, the event that is "A girl is chosen to be the representative", so the $P(A)=\frac{1}{3}$ and $P(\bar{A})=\frac{2}{3}$. Now let be X the random variable that reflex the number of girls chosen.

$P(X=2)=^4C_{2} \cdot\left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^2=6\cdot \frac{1}{9}\cdot \frac{4}{9}=\frac{24}{81}=\frac{8}{27}$

But the book's solution is $\frac{^4C_{2} \cdot 2 \cdot 4^2}{6^4}=\frac{4}{27}$

I think that I'm correct,but can you confirm, please?

Answer 1

What you have written looks sensible. I suspect the book missed a power of $2$ and was aiming to say $$\frac{^4C_{2} \cdot 2^2 \cdot 4^2}{6^4}=\frac{8}{27}.$$

The probabilities for $0,1,2,3,4$ girl representatives are $\dfrac{16}{81},\dfrac{32}{81},\dfrac{24}{81},\dfrac{8}{81},\dfrac{1}{81}$ respectively, summing to $1$, and $\dfrac{24}{81}=\dfrac{8}{27}$.