# The probability that after repeated random drawing from an urn, all balls left in the urn will be red

### Problem

An urn contains $$p$$ red and $$q$$ green balls. Balls are drawn one by one till balls left in the urn are all red. Prove that the probability of this event is $$\dfrac {p}{p+q}$$. Please note that you are told that balls of same colour are left in the urn and you need to find the probability that these remaining balls are red.

### Progress

Okay, so first of all, we note that it is pretty confusing that whether balls of same colour are considered identical or different is not known. I proceeded by assuming the balls are all distinct.

We need a minimum of $$q$$ turns and a maximum of $$p+q-2$$ turns. When we have $$q$$ turns, the $$q$$ green balls are permuted in $$q!$$ ways. Probability = $$\dfrac {q!}{(p+q)P(q)}$$ since in our sample space we need to get arrangement of $$q$$ balls from $$p+q$$ balls.

When we have $$q+1$$ turns, select the green ball which will be in $$q+1$$th place in $${q\choose 1}$$ ways, then select one red ball in $${p\choose 1}$$ ways, then arrange the $$q-1$$ green balls and $$1$$ red ball in $$q!$$ ways, so probability is $$\dfrac { {q\choose 1} { p\choose 1}(q!)}{(p+q)P(q+1)}$$ We go on like this upto $$p+q-2$$ trials and add up all this, but this is not coming out as $$p/(p+q)$$

• I tried many ways, nothing seems to work. But all my methods seem correct to me, at least. – AnonymousMaths Jul 31 '14 at 16:38
• Put some of what you've tried into your question: that'll give us a better sense of your progress and where things may be going wrong. – Semiclassical Jul 31 '14 at 16:40
• Are you familiar with recurrence equations? – Alex Jul 31 '14 at 16:52
• What 'event' exactly? Do you mean the event that it comes that far that after some drawing there are no green balls in the urn anymore while there are still balls in the urn? – drhab Jul 31 '14 at 17:05
• I am familiar with recurrence relations. Can you please show a method to apply such relations here? That is, a complete answer, please? – AnonymousMaths Jul 31 '14 at 17:09

This problem needs a change in notation. $p$ and $q$ look like probabilities and hard to remember which is red and which is green. So, we have a box with $r$ red balls and $g$ green balls. We remove balls one at a time (at random) until there is one ball left. Show that the probability that the last ball in the box is red is $r/(r+g).$

Let the pair $(R_k,G_k)$ be the number of red and green balls remaining in the box after draw $k.$ The pair is an absorbing Markov chain. The absorbing states are $(1,0)$ and $(0,1).$ The initial state is $(R_0,G_0)=(r,g).$ Let $p(i,j)$ be the probability that the last ball in the box is red starting from the state $(i,j).$

Now condition on the result of the first ball drawn: $$p(r,g)=\frac r{r+g}p(r-1,g)+\frac{g}{r+g}p(r,g-1)$$

Clearly, $p(i,0)=1$ for $i\ge 1$ and $p(0,j)=0$ for $j\ge 1.$ So this gives the result for $1$ ball in the box. Now, by induction suppose we know $p(i,j)=i/(i+j)$ when we have $n=i+j$ balls in the box, $1\le i\le n.$ The case $n=1$ holds. Using this on the RHS of the recurrence relation above, $$\frac{r}{r+g}\frac {r-1}{r-1+g}+\frac{g}{r+g}\frac{r}{r+g-1}$$ which equals $\frac{r}{r+g}$ completing the induction by showing the formula is correct for a box with ${n+1}$ balls.

• no!! please see the problem carefully. At least two balls of the same color are remaining, not one ball is remaining. – AnonymousMaths Jul 31 '14 at 22:20

The condition means that if you continue extracting balls, the last one will be red.

There are $\displaystyle p+q\choose\displaystyle p$ ways to order all the balls. If we fix the last one red, we can order the rest of them in $\displaystyle p+q-1 \choose \displaystyle p-1$ ways. The probability is then:

$$\frac{p+q-1 \choose p -1}{p+q \choose p} = \frac{\frac{(p+q-1)!}{(p-1)!q!}}{\frac{(p+q)!}{p!q!}} =\frac{\frac{(p+q-1)!}{(p-1)!}}{\frac{(p+q)·(p+q-1)!}{p·(p-1)!}} = \frac{p}{p+q}$$

• Actually, after clarification, you are taking out balls till you have balls of same colour left. Find the probability that these remaining balls are red. You have to prove that this probability is p/(p+q) – AnonymousMaths Jul 31 '14 at 17:17
• @AnonymousMaths yes, that's what I did. The condition that states that you remove all the balls at random until the remaining ones are red is equivalent as to remove all the balls and that the last one is red. Those two things are equivalent. – Darth Geek Jul 31 '14 at 17:24
• But how is that equivalent? :( Sorry but this apparently means you are implying that 2nd last green, last red = 2nd last red, last red, which is not the case right! – AnonymousMaths Jul 31 '14 at 17:26
• @AnonymousMaths You don't need the penultimate ball to be red or green. If the last ball is red it means that at some point, all the remaining balls in the urn were red. – Darth Geek Jul 31 '14 at 17:29
• This is exactly what I am not getting :( So sorry for that. How does this imply that the remaining balls in the urn were red? :( How can the possibility that there is a mixture of red and green balls be ruled out? – AnonymousMaths Jul 31 '14 at 17:32

I formulate two problems here and give an answer to both of them.

It is up to you to sort out wich of the problems you are dealing with.

Personally I think it is problem 1. This because of its nice answer $\frac{p}{p+q}$.

Draw all $n:=p+q$ balls and let $C_{i}\in\left\{ R,G\right\}$ denote the colour of the $i$th ball drawn.

Problem1:

To find is the probalitity of the event that after drawing less than $n$ balls we land in a situation in which all balls left in the urn are red.

This event is exactly the event $\left\{ C_{n}=R\right\}$ and $P\left\{ C_{n}=R\right\} =\frac{p}{p+q}$

Problem2:

To find is: $P\left\{ C_{n-1}=R=C_{n}\mid C_{n-1}=C_{n}\right\}$ i.e. the probability that the last two balls drawn were red under the condition that they had the same colour.

We find:

$P\left\{ C_{n-1}=R=C_{n}\right\} =P\left\{ C_{n}=R|C_{n-1}=R\right\} P\left\{ C_{n-1}=R\right\} =\frac{p-1}{p+q-1}\frac{p}{p+q}$

$P\left\{ C_{n-1}=G=C_{n}\right\} =P\left\{ C_{n}=G|C_{n-1}=G\right\} P\left\{ C_{n-1}=G\right\} =\frac{q-1}{p+q-1}\frac{q}{p+q}$

$P\left\{ C_{n-1}=C_{n}\right\} =P\left\{ C_{n-1}=R=C_{n}\right\} +P\left\{ C_{n-1}=R=C_{n}\right\} =\frac{\left(p-1\right)p+\left(q-1\right)q}{\left(p+q-1\right)\left(p+q\right)}$

So $P\left\{ C_{n-1}=R=C_{n}\mid C_{n-1}=C_{n}\right\} =\frac{P\left\{ C_{n-1}=R=C_{n}\wedge C_{n-1}=C_{n}\right\} }{P\left\{ C_{n-1}=C_{n}\right\} }=\frac{P\left\{ C_{n-1}=R=C_{n}\right\} }{P\left\{ C_{n-1}=C_{n}\right\} }=\frac{\left(p-1\right)p}{\left(p-1\right)p+\left(q-1\right)q}$

An urn contains $p$ red and $q$ green balls. Balls are drawn one by one till balls left in the urn are all red. Prove that the probability of this event is $p/(p+q)$ . Please note that you are told that balls of same colour are left in the urn and you need to find the probability that these remaining balls are red.

No, just look to the complement.  This event can only not happen if all balls are drawn and the last one turns out to be green.  Of all the $p+q$ balls that could be the last one drawn, just $q$ of them are green.  \begin{align}P(\text{Event}) & = 1-P(\text{Complement}) \\[1ex] & = 1-\frac{q}{p+q} \\[1ex] & = \frac{p}{p+q}\end{align}