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?
Best Answer
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 $d$.