Marble drawing without replacement question There are $7$ marbles in a bag. $4$ white and $3$ black. Marbles are removed from the bag one at a time without replacement. What is the probability that the fifth marble removed is black?
My thought is $\frac37$.
The thinking is like you are trying to label $1$ to $7$ on these balls. The chance that a black marble gets the label $5$ is $\frac37$.

Is my thought correct?

 A: Another way to see this is to imagine that you pull the marbles out of the bag one at a time and arrange them in a line.  Then you add an extra step:  you switch the positions of the first and fifth marbles in the line.
Then the proportion of outcomes in which the first marble is black after switching is the same as the proportion of outcomes in which the first marble is black before switching, because every outcome in which the two marbles are the same color is unaffected by switching, and every outcome in which the two marbles are different colors can be put in one-to-one correspondence with the reverse outcome in which the colors are switched.
Therefore, the original question is the same as asking "what is the probability that the first marble drawn is black?"  And this is obviously $3/7$.
A: The way I like to put it for such problems is:
Black marbles have no preference for position
thus are equally likely to be found at any position with $Pr = \frac37$
A: You want to take the marbles and put them inside boxes labelled $1$ to $7$. You can take the marble for box number $5$ first, hence $\frac{3}{7}$
A: Marbles are removed from the bag one at a time without replacement
Case $1$: If the first $4$ are white then $3$ blacks are available on the fifth draw. i.e $\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{1}{4}(\frac{3}{3})$
or
Case $2$: if you choose one black in the $k$ position such that $k < 5 \frac{3}{7}\frac{4}{6}\frac{3}{5}\frac{2}{4}\frac{2}{3}$
or
Case3: two blacks were chosen prior the fifth draw
i.e $\frac{4}{7}\frac{3}{6}\frac{3}{5}\frac{2}{4}\frac{1}{3}
Hence there are $3$ ways to get a black in the $5$th position of sample space $7 = 3/7$
