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Suppose we have $n$ white and $m$ black balls in a urn. First, randomly withdraw two balls, what is the probability (Call it $P_1$) that they are the same color? Now, suppose a ball is randomly withdrawn and then replaced before second one is drawn, what is the probability (Call it $P_2$) that withdrawn balls are same color? Finally prove that $P_2 > P_1$.

try

For the first situation sample space size is ${m + n \choose 2 }$. Now, in how many ways can we withdraw balls the same color? If both are white, then can do this in ${n \choose 2}$ ways and if both are black can do in ${m \choose 2}$. Thus

$$ P_1 = \frac{ {m \choose 2 } + {n \choose 2} }{ {m+n \choose 2 } } $$

Now, for second situation, two cases. If the first ball drawn is white, then the probability this happens is ${n \choose 1 } / {m+n \choose 1 } = \frac{n}{m+n} $ . For the seecond ball we want it to be white so this can be done in ${n-1 \choose 1 } / {m+n-1 \choose 1 } = \frac{n-1}{m+n-1} $ so for this case we have $\frac{n(n-1) }{(m+n)(m+n-1)}$. Similarly if the first ball drawn is black we obtain probability $ \frac{m(m-1) }{(m+n)(m+n-1)}$.Thus,

$$ P_2 = \frac{ m(m-1) + n(n-1) }{(m+n)(m+n-1) } $$

But, Im stuck in trying to prove $P_2 > P_1$. Is my approach correct?

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  • $\begingroup$ In the second situation , the probability of the second ball being white is the same as that for the first ball because of the replacement $\endgroup$ – WW1 Apr 17 '18 at 4:33
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When working without replacement, we can either select $2$ white balls with probability

$$\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}$$

or $2$ black balls with probability

$$\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$

giving $$P_1=\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}+\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$

which is equivalent to what you have done.

When working with replacement, the probability does not change after the first draw giving

$$P_2=\frac{n}{n+m}\cdot\frac{n}{n+m}+\frac{m}{n+m}\cdot\frac{m}{n+m}$$

It suffices to show that for two positive integers $$\frac{x}{x+y}\gt\frac{x-1}{x+y-1}$$

Can you go from here?

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You are on the right track for $P_1$ , drawing without replacement.

Next, notice that $\dbinom{x}{2}=\dfrac{x\,(x-1)}{2!}$, and so

$$P_1 ~{=\dfrac{\binom m2 +\binom n2}{\binom{m+n}2} \\ = \dfrac{m(m-1)+n(n-1)}{(n+m)(n+m-1)}}$$


However, $P_2$ needs to be rethought. (Notice that you actually evaluated the first situation .)

Rememer, it is drawing with replacement.   That means you won't have one less ball on the subsequent draw (as it's been replaced).

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Since you are drawing the balls with replacement in part 2, you do not need to modify the number of balls being chosen from. The number of ways to choose 1 ball out of n balls is ${n}\choose{1}$. The number of possibilities of the same thing twice is ${{n}\choose{1}}^2$

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