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I am reading probability notes from : On page $8$, there is this question : We have a bag that contains $100$ balls, $50$ of them red and $50$ blue. Select $5$ balls at random. What is the probability that $3$ are blue and $2$ are red?

The answer suggested is $\dfrac{\dbinom{50}{3} \times \dbinom{50}2}{\dbinom{100}5}$.

Here $\dbinom{50}{3}$ means ways to choose $3$ out of $50$.

Does not this answer assumes all red/blue balls are distinct form other red/blue balls, though never mentions in problem.

I think it is not correct.

What I think :

Total number of ways ball can be selected is $2^5$. Ball can be either red or blue. Total favorable events : $\dbinom{5}3$ or $\dbinom{5}2$ i.e. places where red or blue balls can be placed. And thus probability is $$\dfrac{\dbinom{5}3}{2^5}$$

I understand notes can have some bug. Help me to understand which method is correct.

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Notes are correct. Imagine that the balls have numbers on them, written in invisible ink. The $2^5$ events of the analyis you mention are not all equally likely. If we draw, record the colour, replace, draw again, record colour, replace, and so on, then indeed your $2^5$ based analysis becomes correct. – André Nicolas Nov 4 '12 at 5:47
up vote 1 down vote accepted

The answer in the notes is correct. There are $\binom{50}3$ different sets of $3$ blue balls and $\binom{50}2$ different sets of $2$ red balls. Each of the $\binom{50}3$ sets of $3$ blue balls may be paired with any of the $\binom{50}2$ sets of $2$ red balls to form a set of $3$ blue and $2$ red balls, and every set of $3$ blue and $2$ red balls is formed in that way. Thus, there are $\binom{50}3\binom{50}2$ sets of $3$ blue and $2$ red balls. Since there are $\binom{100}5$ different sets of $5$ balls, the probability of drawing a set of $3$ blue and $2$ red balls is


exactly as it says in the notes.

It is true that there are $2^5$ different $5$-term sequences of the colors red and blue, and that $\binom53$ of them have $3$ blue and $2$ red terms, but that’s not what we’re counting. To see what goes wrong here, imagine that the bag contains only $5$ balls of each color. Now it’s clear that there are $5!$ ways to draw the color sequence BBBBB: you must draw the $5$ blue balls in some order. To get the color sequence RBBBB, however, you must first draw one of the $5$ red balls, then one of the $5$ blue balls, then one of the $4$ remaining blue balls, then one of $3$ blue balls left after that, and finally one of the last $2$ blue balls; you can do this in $4\cdot5\cdot4\cdot3\cdot2=480$ different ways. The two color sequences BBBBB and RBBBB are therefore not equally likely; in fact, the latter is four times as likely as the former, and you’re four times as likely to get it when you draw at random.

Another problem with your solution is that the problem isn’t about order: the balls are drawn as a set of $5$ balls, all at once, not as a sequence of $5$ balls.

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Nice explanation. Now I understand difference b/w sequence and bunch ... – Vishal Kumar Nov 4 '12 at 17:25
@Vishal: Thank you; glad it helped. – Brian M. Scott Nov 4 '12 at 17:32

No, the answer is indeed correct. When you're saying there are $2^5$ ways to choose, you're not counting the number of ways to pull balls from the bag. You're counting the number of sequences there are based on colors.

To see why this is incorrect, consider a slightly different scenario of 60 red and 40 blue balls in the bag. There would still be $2^5 = 32$ different sequences of picking out red and blue, but because there are not equal numbers of balls in the bag in the first place, some sequences are obviously more probable than others. Your method of counting doesn't account for this--nor does it account for how picking out a particular color reduces the probability of picking it again on the next pull (which is something that does apply to the original problem as stated).

In essence, it is like we mark or number the balls and count all the different ballswe could pull out of the bag, and only at the end do we enforce that the balls are really indistinguishable (so their orderings in any given group cannot matter). But this is standard practice in combinatorics, and it produces the right results.

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