Choosing two balls when there are not enough balls of a given colour

Question

Suppose we have $4$ balls in a bag, $3$ of them are red and $1$ is blue. We would take two balls at random. What is the probability of taking out:

1. two same colours.
2. two different colours.

I did as following; to have same colour: $$\frac{{3\choose 2}+{1\choose 1}}{{4\choose 2}}\approx0.66$$ to be different colour: $$1-\frac{{3\choose 2}+{1\choose 1}}{{4\choose 2}}\approx0.33$$ Is this correct? I did this intuitively as ${1\choose 2}=0$ means you can not choose two balls from 1 ball left. Am I correct in this justification?

Answer 1

Suppose we have four balls in a bag, three of them are red and one is blue. We draw two balls at random. What is the probability of selecting two balls of the same color?

There are $\binom{4}{2}$ ways to select to select two of the four balls.

The favorable cases are selecting two of the three red balls, which can be done in $\binom{3}{2}$ ways, or two blue balls, which is impossible since there is only one blue ball. Hence, the probability that two balls of the same color are obtained is $$\Pr(\text{two balls of the same color}) = \frac{\dbinom{3}{2}}{\dbinom{4}{2}}$$ If you want to make clear that you considered the possibility of selecting two blue balls, you can write the answer in the form $$\Pr(\text{two balls of the same color}) = \frac{\dbinom{3}{2} + \dbinom{1}{2}}{\dbinom{4}{2}}$$ which is equal to the previous result since $\binom{1}{2} = 0$.

Suppose we have four balls in a bag, three of them are red and one is blue. We draw two balls at random. What is the probability of selecting two balls of different colors?

The denominator is the same as above.

The favorable cases are those in which we select one of the three red balls and the only blue ball, so the probability is $$\Pr(\text{two balls of different colors}) = \frac{\dbinom{3}{1}\dbinom{1}{1}}{\dbinom{4}{2}}$$

Check: There are only two possibilities. Either the selected balls are the same color or they are different colors. Therefore, the two probabilities that we have calculated should add to $1$, as you can verify.

Answer 2

Let $P(a,b)$ be the probability of choosing a ball with colour $a$ followed by a ball of colour $b$. I am going to assume that the two balls are taken out at one after the other without replacement. Then

$P(\text{red}, \text{red}) = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$

$P(\text{red}, \text{blue}) = \frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$

$P(\text{blue}, \text{red}) = \frac{1}{4} \times \frac{3}{3} = \frac{1}{4}$

$P(\text{blue}, \text{blue}) = 0$

I'll let you take it from there.