## Picking Balls

A bag contains four balls – two red, one white and one blue. Someone takes two balls from the bag and tells you that at least one of them is red. What is the probability that the other is red?

This is a simple logic problem which seemed to cause quite a few problems – is red then white the same as white then red, are there two red/red’s or just one.Is the answer 1 in 3 or 1 in 5?

Part of the confusing is that there are two approaches to the problem – looking at the number of *combinations* that you can have with a least one ball being red *or* looking at the number of *ways* you can get two balls with a least one of them being red.

The easiest way to see it is to realise that *order does not matter*. So let us look at the number of *combinations* in which we can take two balls out with one of them being red:

So there are five different combinations. Only one in which both are red – hence the probability of one in five.

In the diagram above there are two *ways* of getting each combination. To make things a bit easier let us assume that the balls are also distinguishable in some other way (e.g. they are different sizes) although this does not really matter.

First we take the case when the first ball picked is red. There are six *ways* of doing this:

With two of these *ways* both balls are red – we could pick the small red ball first then the big red ball or the big red ball then the small red ball. Therefore the probability of picking a red ball if the first ball is red is 2 in 6 which is the same as 1 in 3.

However, the first ball could be white or it could be blue. This gives us another four *ways* of picking the balls so that at least one of them is red:

Therefore we have 6 + 4 = 10 different *ways* of picking the balls so that at least one of them is red. In two of those ways both balls are red. Therefore the probability of the both balls being red if we know that one of them is red is 2 in 10 which is the same as 1 in 5.

Just for completeness let us have a look at the combinations and ways in which neither ball is red.

There is only one combination:

and there are two ways of getting this combination

There are four ways of picking the first ball and three ways of picking the second. Therefore the number of ways of picking two balls is 4 x 3 = 12.

We had 10 ways of picking two balls so that at least one of them is red and another 2 ways where neither of them is red giving a total of 12 ways as expected.

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