0
$\begingroup$

An urn contains $b$ blue balls and $r$ red balls. They are removed at random and not replace. Show the probability that the first red ball drawn is the $(k+1)$th ball drawn equals $\binom{r+b-k-1}{r-1}/\binom{r+b}{b}$.

Because each draw is random and $(k+1)$th draw is red, we already pick $k$ blue balls. The probability for the first $k$ draw is $(b-i)/(r+b-i)$ where $0\leq i\leq k$. When we draw a red ball, the probability is $(r)/(r+b-k)$. I get stuck at this step. 

$\endgroup$

3 Answers 3

1
$\begingroup$

The probability of selecting $k$ from the $b$ balls when selecting $k$ from all $b+r$ balls is $$\require{cancel}\frac{\binom bk}{\binom {b+r}k} = \dfrac{b!}{(b-k)!~\cancel{k!}}\cdot\dfrac{\cancel{k!}~(b+r-k)!}{(b+r)!}$$

The conditional probability that the next ball is red when there are $r$ and $r+b-k$ balls remaining is: $$\dfrac{r}{r+b-k}$$

Multiply and rearrange.


Hints: $x=\frac{x!}{(x-1)!}$ and $x! = x~(x-1)!$

$\endgroup$
0
$\begingroup$

The probability is (taking one ball at a time in this product) $$ \left(\frac{b}{b+r} \right) \left(\frac{b-1}{b+r-1}\right) \cdots \left(\frac{b-(k-1)}{b+r-(k-1)}\right) \left(\frac{r}{b+r-k}\right) $$ The product of the numerators is $$\frac{b!}{(b-k)!} r$$ The product of the denominators is $$\frac{(b+r)!}{(b+r-k-1)!} r$$

So the probability will be $$ \frac{r(b+r-k-1)!b!}{(b-k)!(b+r)!} $$ And we can manipulate this to get: $$ \frac{r(b+r-k-1)!b!}{(b-k)!(b+r)!} = \frac{\frac{r(b+r-k-1)!}{(b-k)!} }{\frac{(b+r)!}{b!}} = \frac{\frac{r(b+r-k-1)!}{r!(b-k)!} }{\frac{(b+r)!}{b!r!}}= \frac{\frac{(b+r-k-1)!}{(r-1)!(b-k)!} }{\binom{b+r}{b}}=\frac{\binom{b+r-k-1}{r-1}}{\binom{b+r}{b}} $$

$\endgroup$
0
$\begingroup$

How can the first draw have a $\frac{b-i}{r+b-i}$? You have not picked any right? You got $b$ blue balls and $r+b$ total balls. The probability that a blue ball might be picked from all balls at first draw is $$\text{total number of blueballs / total number of choices}$$ $$=\frac b{r+b}$$

$\endgroup$
3
  • $\begingroup$ I suspect simple was multiplying for the first $k$ draws: $\prod\limits_{i=0}^{k-1} \frac{b-i}{r+b-i}$ $\endgroup$ Commented Oct 14, 2016 at 11:15
  • $\begingroup$ @user567693 The favoured event was for the first red ball at position $k+1$; meaning no others drawn before that. $\endgroup$ Commented Oct 15, 2016 at 3:45
  • $\begingroup$ Yup, simpler way of writing it. But you might wanna multiply upto k, because we are talking about the probability at k + 1 th draw to be a red ball. Hehewe are the only people talking instincts here, everyone else is going formulas :p(man I'm dumb). $\endgroup$ Commented Oct 15, 2016 at 3:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .