A bag contains $6$ white balls and $8$ blue balls. 
A bag contains $6$ white balls and $8$ blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a) the first is white and the second is blue,
b) both are white
c) One is white and the other is blue.

I tried as:
$6$ white balls + $8$ blue balls = $14$ balls.
So, exhaustive cases$=^{14}C_{2}=48$
a). Favourable cases$=^{6}C_{1}.^{8}C_{1}=48$
Then, Probability$=\dfrac {48}{91}$.
b). Favourable cases$=^{6}C_{2}=15$
Then, Probability$=\dfrac {15}{91}$
The answer for part (a) doesn't match with the answer key. However answer to (b) is correct as per the answer key. Why is it so? I am not able to solve the third part of the question.
 A: In the first problem, you have to take the order of selection into account.  Therefore, the size of your sample space is the $14 \cdot 13$ ways you could select one of the $6 + 8 = 14$ balls, then select one of the remaining $13$ balls.  Since there are six ways to select a white ball on the first draw and eight ways to select a black ball on the second draw, the probability of first selecting a white ball and then selecting a black ball when two balls are drawn is
$$\Pr(\text{selecting white ball, then a black ball}) = \frac{6 \cdot 8}{14 \cdot 13}$$
Another way to see this is to observe that the probability that the first ball selected is white is $6/14$ and that the probability that the second ball selected is black is $8/13$.  Hence,
$$\Pr(\text{selecting white ball, then a black ball}) = \frac{6}{14}\cdot \frac{8}{13}$$ 
In the second problem, what matters is that we select two of the six white balls when we select two of the fourteen balls in the bag, so 
$$\Pr(\text{selecting two white balls}) = \frac{\dbinom{6}{2}}{\dbinom{14}{2}}$$
Had we instead treated the problem as an ordered selection, we would obtain
$$\Pr(\text{selecting two white balls}) = \frac{6}{14} \cdot \frac{5}{13}$$
As you can check, we get the same result in either case.
In the third problem, order does not matter.  Count the favorable cases by selecting one black and one white ball.  Divide by the number of ways of selecting two balls.

 $$\Pr(\text{selecting one black and one white ball}) = \frac{\dbinom{8}{1}\dbinom{6}{1}}{\dbinom{14}{2}}$$

A: In the first problem we have that the probability the first ball is white is $6/14$. Now the balls are $13$ and the blue balls are still $8$, hence the probability the second ball is blue is $8/13$:
$$p_1=\frac{6}{14}\cdot \frac{8}{13}=\frac{24}{91}.$$
A similar approach can be used for the second question:
$$p_2=\frac{6}{14}\cdot \frac{5}{13}=\frac{15}{91}.$$
As regards the third question we have the same probability $p_1$ for the case when first ball is blue and the second is white. So what is $p_3$?
A: For part a) the probability of draw a white ball is $\frac{6}{14}$. After you draw the white ball, there are 13 balls in the bag so the probability of the second to be blue is $\frac{8}{13}$. Now $\frac{6}{14}\cdot \frac{8}{13}=\frac{24}{91}$ is the probability for a). 
For part b) same argument but in this case the probability for the second ball to be white is $\frac{5}{13}$. Therefore the probability for b) is $\frac{6}{14}\cdot \frac{5}{13}=\frac{15}{91}$ 
For part c) now you have two possibilities: first one white and second blue (probability $\frac{24}{91}$) or first one blue and second white (probability $\frac{8}{14} \cdot \frac{6}{13}=\frac{24}{91}$). Therefore, the total probability is the sum $\frac{48}{91}$
A: It seems to me that this problem is easy to reason out without just using "formulas".
A bag contains 6 white balls and 8 blue balls. Two balls are drawn from the bag at random one after another without replacement. Find the probability that:
a). the first is white and the second is blue.
  There are 14 balls and 6 are white.  The probability the first ball drawn is white is 6/14= 3/7.  There are then 13 balls left, 8 of which are blue.  The probability the second ball drawn is 8/13.  The probability of drawing first a white then a blue ball is (3/7)(8/13)= 24/91.
b). both are white
  There are 14 balls and 6 are white.  The probability the first ball drawn is white is 6/14= 3/7.  There are then 13 balls left, 5 of which are blue.  The probability the second ball is also white is 5/13.  The probability of drawing two white balls is (3/7)(5/13)= 15/91.
c). One is white and the other is blue.
We already have that the probability the first is white, the second blue, is 24/91.  Using the same argument, there were initially 14 balls, 8 of them blue.  The probability the first ball is blue is 8/14= 4/7.  There are then 13 balls, 6 of them white.  The probability the second ball is white is 6/13.  The probability the first ball is blue, the second white, is (4/7)(6/13)= 24/91, just as in (a) (that is always true- we have the same numbers, just with numerators switched).  The probability one is white and the other blue, in either order, is 2*(24/91)= 48/91. 
