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

\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.
