Probability: You have 6 colored balls of 3 colors, with 2 of each color. Probability of not drawing a certain pair.

Question

Struggling with how to approach this question. I'm trying to use casework, but it's devolving into a really messy problem very quickly.

You have 6 colored balls of 3 colors, with 2 of each color. For example, 2 red, 2 green, and 2 blue balls. You're drawing for 3 pairs of balls, without replacement. What's the probability that you don't draw a RG/GR pair. For example, a drawing could be as follows:

1. first draw a red, then draw a second red (RR)
2. (from the remaining 4 balls) draw a green, then draw a blue
3. (from the remaining 2 balls) draw a green, and then a blue

Your pairs would be: RR, GB, GB

What's the probability that one of the pairs isn't composed of a green and red ball?

Answer 1

Without loss of generality, let the balls be uniquely numbered as $R_1,R_2,B_1,B_2,G_1,G_2$. This lets us organize our thoughts a bit more freely.

Direct: We know that $R_1$ will need to be paired with something. What is the chance that it gets paired with the other red? What is the chance it gets paired with a blue? If it was the case that $R_1$ got paired with a blue, given that information what is the chance then that $R_2$ gets paired with a blue and not a green?

This line of thought gives us an overall answer of $\frac{1}{5}+\frac{2}{5}\cdot\frac{1}{3} = \frac{1}{3}$

---

Indirect: Let's consider the opposite... what is the probability that we did get at least one RG pair?

What is the chance that $R_1$ gets paired with a green? What is the chance that $R_2$ gets paired with a green? What is the chance that both of these happened simultaneously?

Combining this information by inclusion-exclusion and subtracting away from $1$ gives us the final answer as:

$1-\left(\frac{2}{5}+\frac{2}{5}-\frac{2}{5}\cdot\frac{1}{3}\right) = \frac{1}{3}$

Answer 2

Here's how you would solve this:

1. Find all possible pairs from $R$ (red), $G$ (green), and $B$ (blue).

The possible pairs in the set can be easily found by doing:

$x=\frac{6!}{4!}$

Where $x$ is the total number of arrangements possible ($720$) divided by the total arrangements possible that we are not going to use ($24$), which gives us

$x=30$

Now, we want to find what arrangements of $G$ and $R$ can be made. These are:

$GR$

and

$RG$

Since there are only $2$ combinations that can be made of that one pair, the probability for that one pair is $\frac{2(2)}{30}=\frac{2}{15}$

Let's subtract that from $1$ to get:

$1-\frac{2}{15}=\frac{13}{15}$

Now, for the other $2$ pairs:

Since we know that the balls are not replaced after drawing, it gets easier to solve from here:

1. $\frac{4!}{2!}=\frac{24}{2}=12$ possible combinations.

2. If the first pair is an $RG$ pair, the chance of getting one again is

$\frac{2}{12}=\frac{1}{6}$

Otherwise, the chance of getting an $RG$ pair is still $\frac{1}{6}$, since you are forced to choose $2$ balls again.

Now, for the third pair, the only way you wouldn't pick an $RG$ pair is through these conditions:

\begin{cases} RR\, \text{and then getting}\, GG\, \text{or}\, BB \\ GG\, \text{and then getting}\, RR\, \text{or}\, BB \\ BB\, \text{and then getting}\, GG\, \text{or}\, RR \\ \text{Or getting any possible combination of}\, R/G\, \text{and then}\, B \end{cases}

The chances of this happening are:

$\frac{7}{30}$ multiplied by $\frac{7}{12}$ which gets us $\frac{49}{360}$

The $7$ combinations for pairs $1$ and $2$:

1. $RR$, $GG$, $BB$, $RB$, $BR$, $GB$, and $BG$.

2. $RR$, $GG$, $BB$, $RB$, $BR$, $GB$, and $BG$.

This would leave us no choice but to draw a non-$RG$/$GR$ pair for the third pair.

Edit for context: I got $360$ for the denominator because since there are $30$ possible combinations of the $2$ out of the $6$ balls that you are picking, and since there are $12$ possible combinations of balls left, you multiply $30$ and $12$ to obtain $360$.