# probability of drawing different balls-combinatorics

1) Suppose that a box contains r red balls, w white balls, and b blue balls. Suppose also that balls are drawn from the box one at a time, at random, without replacement. What is the probability that all r red balls will be obtained before any white balls are obtained?

2) Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at random, without replacement.

(a) What is the probability that all r red balls will be obtained before any white balls are obtained?

(b) What is the probability that all r red balls will be obtained before two white balls are obtained?

1) Let's say you have 3 red, 4 white, and 5 white balls. So I assume this question is asking, what is the probability of drawing all 3 red balls?

$\therefore \Pr(3R)= \cfrac{1}{\binom{12}{3}}$

Basically, there's only 1 way to get all reds: RRR, and $\binom{12}{3}$ ways of choosing the balls. So to generalize, I thought

$\therefore \Pr(R)= \cfrac{1}{\binom{r+w+b}{r}}$

But in the solution, $w$ is omitted. Does anyone know why? I don't understand.

2a) Apparently the answer is the same as 1?

2b) I don't understand what (b) means. Can someone please enumerate or give an example? So in my case for part 1, are you calculating the probability of RRRWW?

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For question 1), isn't $w$ simply 0? None is drawn... –  hejseb Sep 16 '13 at 19:31

We have $r$ red, $w$ white, and $b$ blue. The first question you have is about the probability that all the red are drawn before any white are drawn.

Forget about the blues, they are irrelevant. Imagine that we only record whether a drawn ball is red and white. A blue is a non-event. All the red are drawn before any white if and only if the first $r$ recorded draws are red. There are $\binom{r+w}{r}$ equally likely records. Only $1$ of them is all red.

So our probability is $\dfrac{1}{\binom{r+w}{r}}$.

For the "before $2$ white are drawn" the argument is similar. The event we are concerned with can happen in two ways: (i) all the red are drawn first or (ii) there is exactly one white in the first $r$ draws, and then a red (again, we can forget about the blues).

We have already found the probability of (i).

For the probability of (ii), the number of ways to draw $r-1$ red and $1$ white in the first $r$ recorded draws is $\binom{r}{r-1}\binom{w}{1}$. So the probability this has happened is $\frac{rw}{\binom{r+w}{r}}$. Given this happened, there are $w$ balls of significance left, of which one is red. Thus the probability of (ii) is $\frac{rw}{\binom{r+w}{r}}\frac{1}{w}=\frac{r}{\binom{r+w}{r}}$.

Add the probabilities of (i) and (ii).

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1) Since there are $\binom{r+w}{r}$ total ways of drawing the balls and only one of these sequences has all the reds first, the probability of drawing all the red balls before any white ball is $\frac{1}{\binom{r+w}{r}}$.

2) If we want to have all the reds drawn before 2 white balls are drawn, then all the red balls must appear in the first $r+1$ draws; so exactly one white ball appears in the first $r+1$ draws. Therefore the probability of drawing all the reds before 2 whites is given by $\frac{r+1}{\binom{r+w}{r}}$.

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1. Here is how I will solve the 1st problem with r, w and b counts of red, white and blue balls:

Total number of ways all the balls can be ordered (or picked one by one) = (r + w + b) ! ........................(1)

Now, consider the arrangements where 'r' red and 'x' blue balls together appear first, then the first white ball, and then the remaining white and blue balls.

Number of ways to to pick 'x' blue balls is C(b, x). Number of ways to arrange 'r' red balls and 'x' blue balls = (r + x) ! Number of ways to pick one of the white balls as the first white is w. Number of ways to arrange the remaining (w+b-x-1) balls = (w+b-x-1)!

Thus, number of ways to pick the first white after all reds have been picked = Sum (x = 0, 1, ..... b) C(b, x) * (r+x)! * w * (w+b-x-1)! ........................(2)

So, the probability of picking the first white ball after all red balls have been picked is (2) / (1).

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