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Peter and John are designing a game. You take 2 balls (once a time, and without replacement) out 0f 12 balls, where 2 are blue, 3 green, 7 black. Tha gambler needs to pay $10 to be able to play the game, and depending on the colors, he can get prices:

(Sample, Price)

2 blue balls $\rightarrow$ 50

1 green, 1 blue $\rightarrow$ 40

2 green balls $\rightarrow$ 20

1 blue ball, 1 black ball $\rightarrow$ 10

Other sample $\rightarrow$ 0

The random variable X is the gain of a person per game, how to get the probability function of this problem. I mean, you get the probability of each event, and then you assign it to the gain (taking into account that you paid 10 dolars).

Would you use permutations or basic fractions on this problem? Thank you very much. My intuition dicates me to use permutations, but some examples suggest me the other option.

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There are various ways to solve the problem. One could use permutations, or combinations, or simple calculation with fractions. Because we are only picking $2$ balls, "fractions" may feel like the simplest, so we do it first.

Imagine taking the balls out one at a time (it makes no difference).

The probability the first is blue is $\frac{2}{12}$. Given the first was blue, the probability the second is blue is $\frac{1}{11}$. Thus the probability of $2$ blue is $\frac{2}{12}\cdot\frac{1}{11}$. This is $\frac{2}{132}$. Thus the probability that our net gain $X$ (after paying the entry fee) is $40$ is $\frac{2}{132}$.

Next we find the probability of $1$ green, $1$ blue. This can happen in two ways, green and then blue, or blue and then green. The probability the first is green is $\frac{3}{12}$. Given the first was green, the probability the second is blue is $\frac{2}{12}$, so the probability of green then blue is $\frac{3}{12}\cdot\frac{2}{11}$. Do the same for blue and then green, and add. We get $\frac{12}{132}$.

The calculations for the next two probabilities involve very similar calculations, and are left to you.

The probability our gross income is $0$, and therefore that $X=-10$, is $1$ minus the sum of the four "prize" probabilities.

Another way: We use "combinations." Note that in what follows, $\binom{n}{r}$ represents the number of ways of choosing an unordered "hand" of $r$ objects from a set of $n$ distinct objects. In school, this is sometimes denoted by $\text{C}^n_r$, or some variant.

Imagine the balls have ID numbers, to make them distinct. There are $\binom{12}{2}$ ways to choose $2$ balls, all equally likely.

There are $\binom{2}{2}$ ways to choose $2$ blues. So the probability of $2$ blues is $\frac{\binom{2}{2}}{\binom{12}{2}}$.

There are $\binom{3}{1}\binom{2}{1}$ ways to choose $1$ green and $1$ blue. Thus the probability of $1$ green and $1$ blue is $\frac{\binom{3}{1}\binom{2}{1}}{\binom{12}{2}}$.

The next two calculations are similar. And as before, we get the probability of total failure by subtracting the sum of the four prize probabilities from $1$.

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