In this series, I am going to explore some interesting probability puzzles. First, I want to consider a question about selecting balls from an urn.

“An urn contains four colored balls: two orange and two blue. Two balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also orange?”

What an interesting problem. Take a guess at what the answer might be.

At first, I was partial to the idea that the probability would be 1/3. After you have been told the first ball is orange, there is a 1 orange ball left out of the total of 3 remaining balls.

Then, I thought the answer must be 1/6. The probability of drawing an orange on the first selection is 2/4 and the probability of drawing an orange on the second selection is 1/3. Using multiplication, we get 1/6.

Unfortunately, neither of these approaches is correct. In fact, the answer is 1/5. Why 1/5 you might ask? Let’s diagram out the situation.

I labelled the balls: **O1**, **O2**, **B1**, **B2**. We are told that of the 2 balls selected, 1 of the balls is orange. Thus, we must have one of the following situations:

**O1**, **O2**

**O1**, **B1**

**O1**, **B2**

**O2**, **B1**

**O2**, **B2**

As you can see, there are only 5 possible selections where at least 1 of the balls selected is orange. Further, only 1 of these possibilities has 2 orange balls. Thus, the probability that the other ball is also orange is 1/5.

This was very shocking to me. I did not expect such a strange answer from a seemingly simple problem.

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