# Probability of drawing a red ball

An urn has $2$ balls and each ball could be green, red or black. We draw a ball and it was green, then it was returned it to the urn. What is the probability that the next ball is red?

My attempt: I think it is just a probability of $1/4$ because we have 4 colors in total but on the other hand I think i need to use conditional probability:

$$P(R|V)= {P(R\bigcap V)\over P(V)}$$

where $P(V)$ is the probability of drawing a green ball , $P(R)$ is the probability of drawing a red ball but I am not so sure which one would be the correct approach of the problem

I would really appreciate your help :)

• What happens if we draw a black or red ball in the 1st try? Also, why are there 4 colors in total? I am confused. Are there 2 balls or 6 balls? – voldemort Aug 20 '14 at 3:28
• there are just $2$ balls in the urn, one of the balls is green and the other one might be green, red or black – user128422 Aug 20 '14 at 3:30

There are just three colors, not four. Anyway, half of the times you re-draw the first ball, that is green. In the other case, you draw the other ball that may be green, red or black with equal probability, $\frac{1}{3}$. Hence the probability to draw a red ball the second time is $\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}$.

Hint: \eqalign{&P(\hbox{second ball is red})\cr &\qquad=P(\hbox{second ball is red}\,|\,\hbox{second ball drawn is the same ball as the first})\cr &\qquad\qquad\qquad{}\times P(\hbox{second ball drawn is the same ball as the first})\cr &\qquad\qquad{}+P(\hbox{second ball is red}\,|\,\hbox{second ball drawn is the other ball})\cr &\qquad\qquad\qquad{}\times P(\hbox{second ball drawn is the other ball})\cr}

I'm guessing we make the Bayesian assumption that before we draw anything, each of the two balls has an equal chance of being any of the $3$ colors.

Your guess of $1/3$ is incorrect; the intuition is that by drawing a red, you gain some knowledge about the urn, and sort of decreases the chance of drawing a red the second time. This can be made precise in the computation.

Following the conditional probabilities given in David's answer:

\begin{align*}&P(\text{second ball is red})\\ &= P(\text{second ball is red} \mid \text{second ball is same as the first})\cdot P(\text{second ball is the same as the first})\\ &\quad + P(\text{second ball is red} \mid \text{second ball is not the first ball})\cdot P(\text{second ball is not the first ball})\\&= \frac{1}{3}\cdot \frac{1}{2} + 0 \cdot \frac{1}{2} \end{align*}

The "knowledge gained" appears in the term $0 \cdot \frac{1}{2}$; before our knowledge, this term would be $\frac{1}{3} \cdot \frac{1}{2}$ so the whole probability would have been $\frac{1}{3}$, which is the prior guess.

• ...except that $P(\hbox{second ball is red}\,|\,\hbox{second ball drawn is the same ball as the first})$ is zero, as the first ball was green. – David Aug 20 '14 at 3:37
• @David just fixed, thanks – angryavian Aug 20 '14 at 3:37
• Thanks a lot guys I really appreciate it :) – user128422 Aug 20 '14 at 3:41

The probability of drawing any one particular color in the original draw is 1/3. In that situation there are 2 copies of the same 3 color possibility. In that problem, there is a 1/2 probability of choosing one of the 2 sets of 3 colors. The probably is 2 * 1/2 * 1/3 =1/3.

In the second situation there is one green ball and another set of 3 colors. In that situation there is not 2 copies of 3 colors, but 3 different probabilities for the drawing of any color from a single draw: red=1/4; black=1/4 and green=2/4=1/2. Note that the possibility of drawing a green ball must be higher in the 2nd case than it was in the first, becasue of the new inofrmation. That means the probability of drawing the red ball in the second case must be lower according to the new information, that there is now 3 colors + another green--4 possible color outcomes per one draw.

A Baysian update must conclude the same probability. The probability is improved from the original p(grn)=p(red)=p(blk)=1/3 to 1/4 by the new information as:

p(red|grn)=(p(grn|red)*p(red)) / (p(grn|red)*p(red)+p(grn|blk)*p(blk)+p(grn|grn)*p(grn) =(1/4 * 1/3) / (1/4*1/3 + 1/4*1/3 + 1/2*1/3) = 1/4

Scott Stillwell