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My son's teacher and I are having a debate about the correct answer to a question. I have an engineer at hand and he has a mathematician so we both feel well supported. We've also both researched the internet and found answers that we feel support our answers. Since his answer came from this website, I decided to ask this same website. Here is the question from my son's 6th grade quiz:

"A box contains three balls of different colors. The colors are red, white and blue. What is the probability of choosing the same color ball 2 times in a row?" Your choices are: A. 2/3 B. 1/9 C. 1/3 D. 2/27

My son answered B. 1/9. I concurred and so did my husband, the engineer. The test answer sheet also said 1/9. His teacher says it would be 1/3 and he emailed me an explanation from this website about 4 red and 6 white balls (link: Probability of first and second drawn balls of the same color, without replacement).

Honestly, I don't see it. I think that explanation would support 1/9.

We've spent quite a bit of time on this so we really have tried to reconcile on this. Could you help us with this simpler sample set of only 3 balls, one of each color? Note, we have agreed that we can assume that the first ball was replaced since there is no zero% probability in the available answers.

We think that the first pick would 1/3 for any color and the second pick (with the ball replaced in the set) would be 1/3 again. 1/3 x 1/3 = 1/9

He says that the first pick would be a 3/3 (100%) chance of picking any color since the color is not specified, second pick 1/3 so 1/1 x 1/3 = 1/3

I'd really appreciate your help. Thank you.

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    It's $\frac 13$. First choice can be anything, second matches the first with probability $\frac 13$. If you are in doubt, list all $9$ possibilities and count. – 2017-01-19
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    @lulu As per your suggestion, the set of ordered pairs corresponding to sequence of choice are as follows $\mathcal{C}=\{(R,R),(R,W),(R,B),(W,W),(W,R),(W,B),(B,B),(B,W),(B,R)\}$ As you can see, only one ordered pair corresponds with two consecutive choices of the red ball. The answer is 1/9th. Your comment would only be correct in the event of a single choice. The probability of any two concurrent results in two separate experiments is always the probability of the first result TIMES the probability of the second result. – 2017-07-13
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    @rocksNwaves I don't understand. Nothing in the question calls for $(R,R)$...it just says "the same color ball $2$ times in a row." Thus you have the cases $(W,W),(R,R),(B,B)$. – 2017-07-13
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    @lulu I see what you mean. I read the question as pick a color, what are the chances of picking that color twice in a row. You are correct, I added words in my head that weren't there. – 2017-07-13
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    @rocksNwaves No problem, that happens to all of us. – 2017-07-13

3 Answers 3

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Answer is $\dfrac{1}{3}$

Any ball can be taken first and for the second ball to be the same color, there is a probability of $\dfrac{1}{3}$. So, required probability is $\dfrac{1}{3}$

Other argument can be favorable outcomes are $3×1=3$ and total possible outcomes are $3×3=9$ and therefore, probability is $\dfrac{3}{9}=\dfrac{1}{3}$

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The answer is 1/3, since for any pick of the first ball, the probability of picking the same ball next time is 1/3. Another way to look at it, is to check all 9 possible combinations of (first ball, second ball). 3 out of the 9 i.e. blue-blue, red-red, white-white, have the second ball match the first, for a total of 3/9=1/3.

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    If you are only asking about the probability of the second event, and not both events consecutively occurring this would be true. The wording of the question makes it clear that they want to know what the chance of picking the same color twice consecutively is. This answer is incorrect. – 2017-07-13
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Its simple.

Let red ball is picked both times then we have probability = $\frac13 \times \frac13 = \frac19$

Or it can be white both times = $\frac13 \times \frac13 = \frac19$

Or blue both times = $\frac13 \times \frac13 = \frac19$

Probability = $\frac19 + \frac19 + \frac19 = \frac39 = \frac13$