First way


Consider all possible configurations of 5 balls, where three balls are black and the first ball is black.

So, we have two black balls somewhere in two positions from 2 to 5 inclusive, and two white balls in remaining positions.


The number of ways to select two positions from 4 possible positions from 2 to 5 is   =  = 6.

Then two other remaining positions  are filled by black balls, so it does not add new configurations.


Thus we have only 6 distinguished configurations for 2 white and 2 black balls in positions from 2 to 5.


From the other side, the number of all possible distinguishable quadruples of white/black balls in 4 position
from 2 to 5 is  = 16  (since in each of 4 position we may have black or white ball, independently).

Thus, the probability under the problem's question is  P =  = .    ANSWER


         Thus, the problem is just solved in this way.   



                      Second way


Another way to think is to determine how many different 4-letter words of two symbols B and W do exist.

This number is   =  = 6.   (! Coincides with "6" from the solution above !)


In this reasoning, same as in the first reasoning, we use implicitly the fact that in this problem, 
the pool of given balls (10 black and 6 white) works in the same way as if this pool be unlimited.

Therefore, these concrete numbers "10 black balls" and "6 white balls) do not play any role in the solution.



                      Third way

                                                          P(A and B)
Use the formula of the conditional probability  P(A|B) = ------------- .
                                                             P(B)

Here P(B) is the probability to have black ball at the first position and anything in positions from 2 to 5,

which is   = .

The probability of event (A and B) is the ratio   = .
    (Here  is the number of cases (Black or White) in 5 positions).

Thus the final probability in this way is  P =  =  =   again.