Box containing Red and black balls A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is

I have a doubt in this problem . 

There are 10 balls 

Sample Space =  First i pick 1st ball X then 2nd ball X then 3rd ball

So Sample Space = 10 * 9 * 8

Ist trial I pick 1 red ball from 4 red balls = 4C1
2nd trial I pick 1 black ball from 6 black balls = 6C1
3rd trial I pick 1 black ball from 5 black balls = 5C1

a.) Probality = (4C1*6C1*5C1)/(10*9*8)

        OR


b.) I pick 1 red ball from 4 red balls =4C1
    I pick 2 black balls from 6 red balls = 6C2

Probality = (4C1*6C2)/(10*9*8)

Is it a or b ? Pleas elaborate ???
tl;dr


  A box containing 4 red balls and 6 black balls . Three balls are selected randomly from the box one after the another, without replacement . The probality that the selected set has one red ball and two black balls is




Do you think order matters? No. Why? Because if we select $RBB$ or $BRB$ or $BBR$ any of them occuring is favorable, and we are done, you might as well consider order but that would cancel eventually. So total possibilities in three turns:
$$\binom{10}3$$
Ways to do favorable event:
$$\binom62\binom41$$
Probability:
$$P=\frac{\binom62\binom41}{\binom{10}3}=\frac12=50\%$$



Now let me consider order:

Total ways:
$$10*9*8$$
Favourable ways:
$$RBB:\binom41\binom61\binom51\\
BRB:\binom61\binom41\binom51\\
BBR:\binom61\binom51\binom41$$
Or simply no. of ways of arranging $B,R,B$ times "you know":
$$\frac{3!}{2!}6*5*4=6*5*4*3$$
Probability is:
$$\frac{6*5*4*3}{10*9*8}=\frac12=50\%$$
a.) Probality = (4C1*6C1*5C1)/(10*9*8)

   OR

  
  b.) I pick 1 red ball from 4 red balls =4C1 I pick 2 black balls from 6 red balls = 6C2
  
  Probality = (4C1*6C2)/(10*9*8)
  
  Is it a or b ? Pleas elaborate ???


Neither, though (b) is closer.  It is important to observe that the favoured case must always be counted in the same manner as the universal case; and vice versa.

(b) is closer because the numerator is counting the ways to pick 1 of 4 red balls and 2 of 6 black balls.  Which is your favoured case.  Now for the denominator you must count the universal case in the same manner: ways to select any 3 of all 10 balls.

$$\mathbb P = \frac{{^4\mathsf C_1}\cdot{^6\mathsf C_2}}{^{10}\mathsf C_3}$$