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Suppose an urn contains 'w' white balls and 'b' black balls and a ball is drawn from it and is replaced along with 'd' additional balls of the same color. Now a second ball is drawn from it. The probability that the second drawn ball is white is independent of the value of 'd'.

My Attempt:

There will be two cases:

$\text{I}:$ A white ball may be drawn, in that case after the replacement and adding additional balls the total balls will be $=$ b+d+w

Then, the probability of drawing a white ball is $\frac{w + d}{w+b+d}$

$\text{II}:$ A black ball may be drawn, again the total balls have to be $=$ w+b+d

But now, the probability of drawing a white ball is $\frac{w}{w+b+d}$

So, the probability of drawing a white ball on the second draw is $= \frac{2w+d}{w+b+d}$
Which is clearly dependent on d.

But my book says that this statement is true.

Can anybody help me understand this?

Tony
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    You can't just add the probabilities from I and II. Let $A$ be the event the first ball is white and $B$ be the event the second is white. Compute $P(A) P(B|A)+P(A^c)P(B|A^c)$. (You computed $P(B|A)$ and $P(B|A^c)$.) – David Mitra May 07 '19 at 10:57

4 Answers4

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Clearly $$P(X_2=w|X_1=w)=\frac{w+d}{w+b+d}$$ as you say, where $X_i$ is the result of draw $i$.

Also $$P(X_2=w| X_1=b)= \frac{w}{w+b+d}$$ from similar reasoning.

So $$P(X_2=w)=P(X_2=w|X_1=w)P(X_1=w) + P(X_2=w|X_1=b)P(X_1=b)$$

by the law of total probability, and computing this gives

$$\frac{w+d}{b+w+d} \frac{w}{w+b} + \frac{w}{w+b+d}\frac{b}{b+w} = \frac{(w+d)w + wb}{(w+b)(w+b+d)} = \frac{w}{w+b}$$ as the common factor $w+b+d$ cancels out.

So $P(X_2=w)$ indeed does not depend on $d$.

Christian Blatter
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Henno Brandsma
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Imagine that the $w$ white balls are all slightly different shades of white, and same for the $b$ many black balls. So there are $w+b$ balls, each of a different shade.

Now by symmetry, the $w+b$ many possible shades for the 2nd ball are all equally likely, since nothing in he process favors one of the shades over another (having started with the same number [i.e. 1] of each shade).

So the the probability that the 2nd ball is white is

(number of white shades)/(total number of shades) = $w/(w+b)$

Ned
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  • +1 really neat argument via "color blindness" or "scanning a color photo using a bad black-and-white scanner" :D – antkam May 07 '19 at 15:43
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    @antkam And the symmetry argument works without modification for iterated draws and/or random numbers of added balls, as long as the initial draw is uniform and number of added balls at each step is independent of the colors chosen so far. – Ned May 08 '19 at 12:35
  • yeah, it took me a bit of time to grasp your solution (and to realize that yours and mine are basically the same argument). – antkam May 08 '19 at 12:37
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Here's an intuitive (as opposed to algebraic) proof.

The key to this question is that the same number, $d$, would be added in either case. So before the second draw, there are $w+b+d$ balls in there. Imagine you can distinguish the "original" $w+b$ balls vs the "newly added" $d$ balls.

Conditioned on you drawing an original ball, $Prob(white|original) = {w \over w+b}$

Conditioned on you drawing a new ball, $Prob(white|new) = {w \over w+b}$ because all new balls are the same color and that color is white iff the first drawn ball is white, which happens with prob ${w \over w+b}$.

The overall probability is a linear combination of the two terms above, with coefficients depending on $d$ and summing to $1$, but since the two terms are equal, it doesn't matter what $d$ is.

BONUS 1: This proof also shows (inductively) that you can extend the process to adding more balls matching the $2$nd draw, then making a $3$rd draw, then adding even more balls matching the $3$rd draw, then making a $4$th draw, etc., and the unconditional prob of drawing white at the $n$th draw will always be ${w \over w+b}$.

BONUS 2: With a bit more work, this proof also shows that $d$ can be a random variable (e.g. how about $d \sim Poisson(\lambda=13)$), as long as it is independent of the first draw result.

antkam
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enter image description here

Let us think of a basic Probability concept. Suppose, the original contents of the urn are w white balls and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. Then the probability that the second ball drawn is white is $\frac{w}{w+b}$, i.e. that this probability does not depend on d. Again, the probability that the second ball drawn is Black is $\frac{b}{w+b}$, i.e. this probability also does not depend on d. Please go through the diagram, where, the required probability does not depend on additional balls denoted by d has been established. Therefore, the probability that the 100th or nth ball you pick is black is $\frac{b}{w+b}$ and White is $\frac{w}{w+b}$

Soudipta Dutta
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