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?
 A: 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
A: This is a case of conditional probability. We're given that the third marble is black, so we have to examine the different possibilities and probabilities of how we could get to that point. The different colorings for the first three marbles given that the third one is black are:


*

*WWB

*WBB

*BWB

*BBB


Within each of these, the probability we arrive at for drawing a black marble on the third draw will be $\frac{20-b}{38}$ where $b$ is the number of black marbles already drawn. The probabilities for drawing the third marble to be black will then be: 


*

*WWB - $\frac{20}{38}$

*WBB - $\frac{19}{38}$

*BWB - $\frac{19}{38}$

*BBB - $\frac{18}{38}$
The probability of the first marble being white will then be $\frac{20+19}{20+19+19+18} = \boxed{\frac{39}{76}}$
