I am reading probability notes from : http://www.stat.berkeley.edu/~aldous/134/gravner.pdf
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.
Best Answer
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
$$\frac{\binom{50}3\binom{50}2}{\binom{100}5}\;,$$
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.