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I have 3 white balls and 4 black balls in a box to be taken without replacement. Let $P(W_i)$ and $P(B_j)$ denote the probability of drawing a white ball at the i-th withdrawal and probability of drawing a black ball at a j-th withdrawal respectively.

I want to find simply $P(W_1)$, $P(B_2)$, $P(W_3)$. The answers are $3/7$, $4/7$ and $3/7$ respectively. But I don't understand why is this the case.

I wanted to use this approach for finding some $P(B_2)$ for example:

$$P(B_2) = \frac{{3\choose 1} {4\choose 1} + {4\choose 1} {3\choose 1}}{{7\choose 2}}$$

To account for both cases where we first get a white ball and then a black ball (WB case) and then the second case where we can get a black ball then another black ball (BB case). But the probability is greater than 1 which is definitely false. What am I missing out?

The explanation for the answer is that it makes no difference to calling which ball is the "i-th" withdrawal and so on, since we can choose any number of balls initially, and from this subset of balls we shuffle them and can get any arbitrary ordering. Is this a valid reasoning? Why?

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    $\begingroup$ The probability of selecting a white ball, then a black ball is $\frac{3}{7} \cdot \frac{4}{6}$ or $\frac{1}{2} \cdot \frac{\binom{3}{1}\binom{4}{1}}{\binom{7}{2}}$. The factor of $1/2$ is necessary since there are two possible orders in which a white ball and a black ball could be selected. The probability of selecting two black balls is $\frac{4}{7} \cdot \frac{3}{6}$ or $\frac{\binom{4}{2}}{\binom{7}{2}}$. Notice that $\frac{3}{7} \cdot \frac{4}{6} + \frac{4}{7} \cdot \frac{3}{6} = \frac{4}{7}$. $\endgroup$ – N. F. Taussig Jun 9 '18 at 13:19
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By symmetry each individual ball has the same probability ${1\over7}$ to be the second ball drawn. Since there are four black balls, and at most one of them can be the second ball drawn, we can just add the probabilities and obtain ${4\over7}$ as probability of the event $B_2$.

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