Probability of trials without replacement using conditional probability 
A bag contains $5$ red marbles and $3$ black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?

Attempt:1
$$
E:\text{first marble is red}\\
F:\text{at least one of the marbles drawn be black}\\
F':\text{none of the marbles drawn be black}\\
n(S)=8*7*6\\
n(F'\cap E)=n(R_1R_2R_3)=5*4*3\\
n(E)=n(\{(R_1R_2R_3),(R_1B_2R_3),(R_1R_2B_3),(R_1B_2B_3)\})\\=5*4*3+5*3*4+5*4*3+5*3*2=60+60+60+30=210\\
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})
=P(F|E)\\&=1-P(F'|E)=1-\frac{60}{210}=1-\frac{2}{7}=\color{red}{\frac{5}{7}}
\end{align}
$$
Attempt:2
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})
=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)\\&=1-P(R_1)P(R_2|R_1)P(R_3|R_1R_2)=1-\frac{5}{8}.\frac{\frac{5*4}{8*7}}{\frac{5}{8}}.\frac{\frac{5*4*3}{8*7*6}}{\frac{5*4}{8*7}}=1-\frac{5}{28}\\&=\color{blue}{\frac{23}{28}}
\end{align}
$$
OR
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)=1-\frac{5*4*3}{8*7*6}=1-\frac{5}{28}=\color{blue}{\frac{23}{28}}
\end{align}
$$
Attempt:3
$$
\begin{align}
&P(\text{at least 1 of the 3 marbles black|1$^\text{st}$ marble red})=P(F|E)\\&=P(R_1B_2R_3)+P(R_1R_2B_3)+P(R_1B_2B_3)\\&=\frac{5*3*4}{8*7*6}+\frac{5*4*3}{8*7*6}+\frac{5*3*2}{8*7*6}=\frac{10+10+5}{4*7*2}=\color{green}{\frac{25}{56}}
\end{align}
$$
The solution of the above problem is stated to be $\frac{25}{56}$ as in my Attempt 3, but why am I not getting the right solution in Attempt $1$ and Attempt $2$ ?
 A: I use the elementary events for the answer. Firstly I use your definition of events:
$E:\text{first marble is red}\\
F:\text{at least one of the marbles drawn be black}\\$
You have to find the probability that you draw at least 3 black marbles given the first marble is red. This is $P(F|E)$
In total we have these 8 cases:
$$rrr, rrb, rbr, brr, bbr, brb, rbb, bbb$$
Now we list the cases where  at least 1 black marble is drawn and the first marble is red.
$$rrb, rbr, rbb$$  
The probability is $$P(F\cap E)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{4}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{2}{7}=\frac{25}{56}$$ 
Now we calculate P(A). The events are $rrr, rrb, rbr,  rbb$ Therefore 
$$P(E)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{4}{6}+\frac{5}{8}\cdot \frac{3}{7}\cdot \frac{2}{6}=\frac{5}{8}$$
Now we apply the  Bayes theorem
$$P(F|E)=\frac{P(F\cap E)}{P(E)}=\frac{\frac{25}{56}}{\frac{5}{8}}=\frac{25}{56}\cdot \frac{8}{5}=\frac{5}{56}\cdot \frac{8}{1}=\frac{5}{7}\cdot \frac{1}{1}=\boxed{\frac57}$$


$$ \begin{align} &P(\text{at least 1 of the 3 marbles
 black|1$^\text{st}$ marble
 red})=P(F|E)\\&=1-P(F'|E)=1-P(R_1R_2R_3)=1-\frac{5*4*3}{8*7*6}=1-\frac{5}{28}=\color{blue}{\frac{23}{28}}
 \end{align} $$

Here you miss the Bayes theorem as well.
$$P(F'|E)=\frac{P(F'\cap E)}{P(E)}$$
$P(F'\cap E)=\frac{5\cdot 4\cdot 3}{8\cdot 7\cdot 6}$
And $P(E)=\frac58$ Thus 
$$P(F|E)=1-P(F'|E)=1-\frac{P(F'\cap E)}{P(E)}=1-\frac{\frac{5\cdot 4\cdot 3}{8\cdot 7\cdot 6}}{\frac58}=1-\frac27=\boxed{\frac57}$$
Many ways lead to Rome.
