Probability of first ball is red or second ball is red 
A bag contains $2$ red and $6$ black balls. Two balls are drawn in succession without replacement from a bag. Then find the probability that the first  ball is red or the second ball is red.

What I have tried: Let $A: $ first ball is red and $B: $ second ball is red.
So $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
Here $\displaystyle P(A)=\frac{2}{6}\times 1=\frac{1}{3}$
And $\displaystyle P(A\cap B)=\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$
I did not understand. How do I solve $P(B)$? Help me please
 A: \begin{align}
P(B) &= P(A \cap B) + P(A^c \cap B)\\
&= P(A \cap B) + \frac{6}{8} \cdot\frac{2}{7}
\end{align}
Also, $P(A) = \frac{2}{2+6}$.
We do not have to compute $P(A \cap B)$ as it will cancel out.
A: As long as you don't know what the first ball was, the second ball is just a uniformly random ball so the probability of $B$ is still $\frac13$.
If you don't find this convincing, you can check by considering the two possibilities: if $B$ occurs either both balls are red (probability $\frac1{15}$ as you already have) or the first ball is black and the second red (probability $\frac46\times \frac25=\frac{4}{15}$) so the total is $\frac5{15}$.
[edit: as Siong Thye Goh points out, the calculation in the question of $1/3$ is incorrect because there are $6$ black balls, not $6$ total balls.]
A: They are asking you to calculate the probability of the following cases:
$$\{RB,BR,RR\}$$
thus simply
$$1-\frac{6}{8}\times \frac{5}{7}+=\frac{13}{28}$$
that is the complement of $\mathbb{P}\{BB\}$
A: Let A : First Ball is Red
B: Second Ball is Red.
You need to find:
$ P(A) P(B|A) + P(A)P(B^c | A) + P(A^c)P(B | A^c)$
These are respectively the probabilities for RR,RB,BR
A: 

A bag contains 2 red and 6 black balls.

I did not understand. How do I solve P(B)?

Each individual ball has the same chance to be the first ball drawn.   Two among those eight are red.
Each individual ball has the same chance to be the second ball drawn.   Two among those eight are red.
The probability for obtaining two from two red balls when selecting two from eight balls is $\tbinom 22/\tbinom82$.
$\mathsf P(A)=1/4, \mathsf P(B)=1/4, \mathsf P(A\cap B)=1/28$ $$\mathsf P(A\cup B)=\dfrac{13}{28}$$

The probability for obtaining two from six black balls when selecting two from eight balls is $\tbinom 62/\tbinom 82$
$$\mathsf P((A\cup B)^\complement)=\dfrac{15}{28}$$
