<s>[INST] probability of selection without replacement An urn contains 20 black marbles and 20 white marbles. Three marbles are chosen without replacement. What is the probability that the first marble is white given that the third marble was black?
 [/INST] You could look at the various probabilities for the eight possibilities for the first three marbles, 
but a quicker way is to use symmetry (each marble can be in any position) and say this is the same as the probability that the second marble is white given that the first marble was black, and that is $$\frac{20}{39}$$

coronermclarson came up with a different answer.  I believe the long-winded answer is to look at the probabilities of the possible patterns for the first three marbles:


*

*$BBB$: $\frac{20}{40}\times \frac{19}{39} \times \frac{18}{38} = \frac{9}{78}$

*$BWB$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$

*$WBB$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$

*$WWB$: $\frac{20}{40}\times \frac{19}{39} \times \frac{20}{38} = \frac{10}{78}$

*$WWW$: $\frac{20}{40}\times \frac{19}{39} \times \frac{18}{38} = \frac{9}{78}$

*$WBW$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$

*$BWW$: $\frac{20}{40}\times \frac{20}{39} \times \frac{19}{38} = \frac{10}{78}$

*$BBW$: $\frac{20}{40}\times \frac{19}{39} \times \frac{20}{38} = \frac{10}{78}$
which add up to $1$, as they should
We are only interested in the first four of these which have the third black, making the probability that the first marble is white given that the third marble was black$$\dfrac{\frac{10}{78}+\frac{10}{78}}{\frac{9}{78}+\frac{10}{78}+\frac{10}{78}+\frac{10}{78}} = \dfrac{10+10}{9+10+10+10} = \dfrac{20}{39}$$ as before
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