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In a bag there are $3$ black balls and $9$ white balls. All the balls are drawn one by one without replacement. What is the probability that the first and the second balls have different colors?

In this kind of problem, I always assume that the balls of the same color (and shape) are indistinguishable between them. So, in this problem, I applied this strategy:

Possible cases: $\frac{12!}{3! \cdot 9!}$

Favorable cases:

We can have BW or WB. For each of these situations the other $10$ balls can be chosen in $\frac{10!}{2! \cdot 8!}$ ways.

In the end, the probability is equal to $\frac{9}{22}$. The book's solution is the same; however, the book's strategy considers each ball as individually unique, despite the color or the shape.

What would you do? Thanks.

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A more interesting problem: What is the probability that the $4$-th and $7$-th balls are of different colours? (The answer is the same.) About unique versus non-unique, the first can be safer. There are situations where if we use non-unique, it is easier wrongly to assume that two events are equally likely when they are not. A perhaps too simple example: We toss two pennies. One could argue that there are $3$ possibilities, two heads, two tails, and mixed, and therefore (???) the probability of two heads is $1/3$. Viewing the coins as distinct saves us from that. – André Nicolas Feb 21 '12 at 22:27
up vote 4 down vote accepted

Like you, I would assume that balls of the same color are indistinguishable. I’d work the problem very differently, however. The probability of drawing a black ball and then a white ball on the first two draws is $$\frac3{12}\cdot\frac9{11}=\frac14\cdot\frac9{11}\;,$$ and the probability of drawing a white ball and then a black ball is $$\frac9{12}\cdot\frac3{11}=\frac34\cdot\frac3{11}\;,$$ so the probability of drawing balls of different colors is $$\frac14\cdot\frac9{11}+\frac34\cdot\frac3{11}=\frac{18}{44}=\frac9{22}\;.$$

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The Law of Total Probabilities to the rescue once more. – Brent Worden Feb 22 '12 at 0:20

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