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.
Best Answer
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.