Example - Drawing balls from a box without replacement
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:
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:
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:
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:
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: