# Example - Drawing balls from a box without replacement¶

Consider the following information I:

We are given a box with 10 balls 6 of which are red and 4 of which are blue. The box is sufficiently mixed so that when we get a ball from it, we don’t know which one we pick. When we take a ball out of the box, we do not put it back.

Fig. 4 A box with 10 balls.

Now, let’s say that we draw the first ball. Let $$B_1$$ be the logical proposition:

The first ball we draw is blue.

What is the probability of $$B_1$$? Our intuition tells us to set:

$p(B_1|I) = \frac{4}{10} = \frac{2}{5}.$

This is known as the principle of insufficient reason. We can now use the obvious rule to find the probability of drawing a red ball, i.e., of $$\neg B_1$$. Of course, $$\neg B_1$$ is just the sentence:

The first ball we draw is red.

So, let’s also call $$\neg B_1$$ with the name $$R_1$$. It is:

$p(R_1|I) = p(\neg B_1|I) = 1 - p(B_1|I) = 1 - \frac{2}{5} = \frac{3}{5}.$

Great! Let’s continue with drawing a second ball. Consider the sentence $$R_2$$:

The second ball we draw is red.

What is the probability of $$R_2$$ given that $$B_1$$ is true? We just need to use common sense to find this probability:

• We had 10 balls, 6 red and 4 blue.

• Since $$B_1$$ is true (the first ball was blue), we now have 6 red and 3 blue balls.

• Therefore, the probability that we draw a red ball next is:

$p(R_2|B_1,I) = \frac{6}{9} = \frac{2}{3}.$

Similarly, we can find the probability that we draw a red ball in the second draw given that we drew a red ball in the first draw:

• We had 10 balls, 6 red and 4 blue.

• Since $$R_1$$ is true (the first ball is red), we now have 5 red and 4 blue balls.

• Therefore, the probability that we draw a red ball next is:

$p(R_2|R_1,I) = \frac{5}{9}.$

All, this was easy. Let’s find the probability that we draw a blue ball in the first draw and a red ball in the second draw. This time, we have to use the product rule:

$p(B_1, R_2|I) = p(R_2|B_1,I) p(B_1|I) = \frac{2}{3}\frac{2}{5} = \frac{4}{15}.$