Basic Probability question- 
Given a bag that contains $6$ Red and $4$ blue balls, what is the probability that the second ball drawn is red?

I am well aware of the standard way to calculate this
$$\mathbb P(\text{first ball is red}) \cdot \mathbb P(\text{second ball is red}) + \mathbb P(\text{first ball is blue}) \cdot\mathbb P(\text{second ball is red})$$
However, when I look at this from my understanding of conditional probability, my understanding fails.
An intuitive approach would be, $$\begin{align}\mathbb P(\text{second ball is red}) &= \mathbb P(\text{second ball is red}\mid\text{first ball is red}) \\ &+ \mathbb P(\text{second ball is red}\mid\text{first ball is blue})\end{align}$$
which turns out to be $\dfrac59 + \dfrac69$ which is obviously incorrect.
Can someone please help explain why the above approach is incorrect?
 A: Let's denote the drawn balls as $X,Y \in \{R,B\}$. The correct expression in your approach should be 
$$
\mathbb{P}[X_2=R|X_1=R] \cdot \mathbb{P}[X_1=R]
+ \mathbb{P}[X_2=R|X_1=B] \cdot \mathbb{P}[X_1=B]
$$
which is the same as your original calculation since $$\mathbb{P}[A|B] \cdot \mathbb{P}[B] = \mathbb{P}[A\cap B]$$ by the definition of conditional probability.
A: Summarizing the comments, both of your formulas are wrong.
Let's look at the second formula first:
$$\begin{align}\mathbb P(\text{second ball is red}) = 
{}& \color{red}{\mathbb P(\text{second ball is red}\mid\text{first ball is red}) }\\
 &+ \color{red}{\mathbb P(\text{second ball is red}\mid\text{first ball is blue})}.\end{align}$$
I don't know what makes this "intuitive"; you would first have to have an intuitive understanding of what a conditional probability is in order to intuitively derive a formula using conditional probabilities.
I would guess that your intuition about conditional probability is misinformed.
If you think about the possibilities for the first ball,
you really have to consider the possibility that it might (or might not!) be red and might (or might not!) be blue.
The "second ball is red" event then partitions into two disjoint sub-events:


*

*The first ball is red and the second ball is red.

*The first ball is blue and the second ball is red.


So one correct formula is,
$$\begin{align}\mathbb P(\text{second ball is red}) 
= {}& \mathbb P(\text{second ball is red }\mathbf{and}\text{ first ball is red}) \\
&+ \mathbb P(\text{second ball is red }\mathbf{and}\text{ first ball is blue}).\end{align} \tag1$$
You can rewrite this using conditional probabilities. Use the fact that
$P(A \cap B) = P(B \mid A) \cdot P(A).$
For example,
\begin{multline}
\mathbb P(\text{second ball is red }\mathbf{and}\text{ first ball is red}) =\\
\mathbb P(\text{second ball is red}\mid\text{first ball is red}) \cdot
\mathbb P(\text{first ball is red}).
\end{multline}
Applying this to the correct formula $(1)$, we get another correct formula,
$$
\begin{align}\mathbb P(\text{second ball is red}) =\qquad& \\
 \mathbb P(\text{second ball is red}\mid{}&\text{first ball is red}) {}\cdot{} \mathbb P(\text{first ball is red})\\
 {}+ \mathbb P(\text{second ball is red}&\mid\text{first ball is blue}) \cdot
\mathbb P(\text{first ball is blue}).\end{align}\tag2
$$
So there are two ways to correct your incorrect "intuitive" formula.
One way is to use "and" instead of "given that".
The other way is to multiply each of the conditional probabilities by the probability of its "given" part.
In the first formula you wrote,
$$\mathbb P(\text{first ball is red}) \cdot \color{red}{\mathbb P(\text{second ball is red})} + \mathbb P(\text{first ball is blue}) \cdot\color{red}{\mathbb P(\text{second ball is red})},$$
since both terms have the same factor,
$\color{red}{\mathbb P(\text{second ball is red})},$
you can pull this factor out using the distributive property:
$$(\mathbb P(\text{first ball is red})
 + \mathbb P(\text{first ball is blue}))
 \cdot\color{red}{\mathbb P(\text{second ball is red})},$$
and since the first ball must be either red or blue,
$\mathbb P(\text{first ball is red})
 + \mathbb P(\text{first ball is blue}) = 1$
and this formula just comes down to $\mathbb P(\text{second ball is red}).$
So that's kind of useless; it tells us that if we already know
$\mathbb P(\text{second ball is red})$ we can find the value of
$\mathbb P(\text{second ball is red}).$
But if you do the individual multiplication
$\mathbb P(\text{first ball is red}) \cdot \color{red}{\mathbb P(\text{second ball is red})},$
you get a meaningless value.
The result of that multiplication is not
$\mathbb P(\text{first ball is red and second ball is red}),$
because the event "second ball is red" is not independent
of the event "first ball is red".
One way that is standard in a situation like this is to multiply
$\mathbb P(\text{first ball is red})$ by
$\mathbb P(\text{second ball is red}\mid\text{first ball is red})$
instead of $\mathbb P(\text{second ball is red})$
and multiply $\mathbb P(\text{first ball is blue})$ by
$\mathbb P(\text{second ball is red}\mid\text{first ball is blue})$
instead of $\mathbb P(\text{second ball is red})$.
Of course, if you do that you end up with Equation $(2)$ as shown above.
