# Probability of drawing the same color marble twice in a bag of $10$ red, $10$ orange, $10$ green.

The question is worded as follows.

A bag of marbles contains $10$ red marbles, $10$ orange marbles, and $10$ green marbles. You randomly take one marble out of the bag for yourself, and without replacement you then randomly take another marble out of the bag to give to your roommate. What is the probability that your roommate has the same color of marble as you do?"

The answer says that it is $10/30 \times 9/29 = 0.103$. I understand that this question is testing your understanding of "dependent events," where the occurrence of one event affects the likelihood of another event in the sequence.

What I don't understand is why it isn't simply $9/29$. I'll explain my thinking, and hopefully someone can point to where I'm wrong.

On the first draw, there are $30$ total possible marbles that can be matched by your roommate. Draw $1$: P{drawing a matchable marble} = $30/30$ On the second draw, there are $9$ marbles of the same color remaining out of 29. Draw $2$: P{drawing the same color} = $9/29$.

$$30/30 \times 9/29 = 9/29.$$

It seems to me that the answer to this problem matches the question: "what is the probability you both draw a green marble?" However, my professor told me that wasn't the case, and that the answer is correct.

Can someone help me? Thanks :)

EDIT Turns out the problem was incorrect, the answer actually is 9/29! Thank you guys for your help :)

• Funny, I would answer the same as you.... And I have no clue as to why the answer should be $10/30 \cdot 9/29$ – RGS Jan 24 '17 at 0:28
• I agree, the answer should be $\frac 9{29}$. The given answer makes sense if you specify the color in advance. That is, if you asked "what is the probability that both you and your roommate get red marbles", say. – lulu Jan 24 '17 at 0:30
• I am guessing that the solution writer fell for the same trick as people fall for in this classic problem – suomynonA Jan 24 '17 at 0:32

Well, we will do the following:

First, fix a colour. This is done in one of three ways.

Then, we are now going to calculate the probability that both you and your friend pick that colour ball out of the bag.

For you, that probability is $$\frac {10}{30}=\frac 13$$, and for your friend, it is $$\frac 9{29}$$. So the answer is $$3 \times \frac{1}{3} \times \frac {9}{29} = \frac{9}{29}$$, which is, your answer.

So you are correct. I find it honestly atrocious (although everybody is entitled to a mistake) to see that after numerous reviews, these mistakes are present in textbooks. It is elementary probability, so if you would have believed both the book and your professor, you would have gone down the wrong path. So, I should say, it was good you asked this question.

EDIT: As you may already have noticed, the textbook's answer would have been correct if you had fixed the colour of the ball before drawing. But then, that's no excuse for such a terrible mistake.

If your teacher still does not believe, perhaps you can try a simulation. This does not prove that you are right, but it is illustrative. The following code will work in R, but you could do something similar with other languages

# Make a vector showing the colors of marbles
color=rep(c(1,2,3), each=10)

# Decide how many times your want to draw out two marbles
nSamples = 10000
# Make a vector to hold the results
Match = rep(NA, nSamples)

# Draw out two marbles without replacement nSamples times
for(i in 1:nSamples){
sampleTemp = sample(color, 2, replace=FALSE)
Match[i] = sampleTemp == sampleTemp
}

# Check the fraction of samples in which the colors match
sum(Match)/nSamples


That is nowhere near (10/30)*(9/29) but will almost always be very close to 9/29.