I'm new to probability theory. I'm trying to do some exercices as the one suggested below but I've come across a solution that I didn't quite understand. Can you, guys, help me figure this out?
A bag contains 2n balls ($n\neq 0$), of which n are white and n are black. All the balls are indistinguishable by touch.
A game consists of drawing a ball from the bag, noting its color, and putting it back in the bag, then drawing another ball from the same bag and noting its color as well.
The rules of the game are as follows:
If both balls drawn are white, you win 20 points (+).
If both balls drawn are black, you lose 20 points (-).
If the two balls drawn are of different colors, the gain is zero (0).
The game is repeated 5 times.
Calculate the probability of winning 40 points.
I went like this:
Let $A$ be the event of obtaining 2 white balls, $B$ obtaining 2 black balls, $C$ obtaining 2 different colors and $E$ winning 40 points at the end.
We have: $P(A)=P(B)=\frac{1}{4}$ and $P(C)=\frac{1}{2}$.
There is two possible cases of gaining 40 points: (+ + 0 0 0) and ( + + + - 0) with all of the different combinations of those two cases. So, $$P = C^2_5 P(A)^2 P(C)^3 + \cfrac{5!}{3!1!1!} P(A)^3 P(C) P(B)$$
However, the solution suggested was assigning a random variable $X$ that is equal to the number of times of gaining 20 points needed to gain $k$ points. They said that this random variable follows a binomial distribution $\mathcal{B}(5,P(A))$. Then, $$P = C^2_5 P(A)^2 (1-P(A))^3$$
But I don't quite understand the solution. I think it is wrong because it takes in consideration the case ( + + - - 0) while the final gain will be 0 points. Am I wrong?