Understanding conditional probability. Consider the problem $(b)$ (from Hwi Hsu): 

This problem can be visually represented in either of the three very closely related ways:
 


Why is $\frac{1}{3}$ only the correct answer? 

1) Probability of rolling doubles given no other information. $P(A)$
$\begin{array}(   \color{blue}{(1,1)} & (1,2)& (1,3)&(1,4)&(1,5)&(1,6)\\
  (2,1) & \color{blue}{(2,2)}& (2,3)&(2,4)&(2,5)&(2,6)\\
  (3,1) & (3,2)& \color{blue}{(3,3)}&(3,4)&(3,5)&(3,6)\\
  (4,1) & (4,2)& (4,3)&\color{blue}{(4,4)}&(4,5)&(4,6)\\
  (5,1) & (5,2)& (5,3)&(5,4)&\color{blue}{(5,5)}&(5,6)\\
  (6,1) & (6,2)& (6,3)&(6,4)&(6,5)&\color{blue}{(6,6)}\\
\end{array}$
Of all the $36$ possible (black and blue), $6$ of them are doubles (blue) so probability is $\frac {6}{36} = \frac 16$.
2) Probability of rolling doubles given that the sum is at most $3$. $P(A|B)$
$\begin{array}(   \color{blue}{(1,1)} & (1,2)& \color{red}{(1,3)}&\color{red}{(1,4)}&\color{red}{(1,5)}&\color{red}{(1,6)}\\
  (2,1) & \color{red}{(2,2)}& \color{red}{(2,3)}&\color{red}{(2,4)}&\color{red}{(2,5)}&\color{red}{(2,6)}\\
  \color{red}{(3,1)} & \color{red}{(3,2)}& \color{red}{(3,3)}&\color{red}{(3,4)}&\color{red}{(3,5)}&\color{red}{(3,6)}\\
 \color{red}{ (4,1)} & \color{red}{(4,2)}& \color{red}{(4,3)}&\color{red}{(4,4)}&\color{red}{(4,5)}&\color{red}{(4,6)}\\
\color{red}{  (5,1)} & \color{red}{(5,2)}& \color{red}{(5,3)}&\color{red}{(5,4)}&\color{red}{(5,5)}&\color{red}{(5,6)}\\
 \color{red}{ (6,1)} &\color{red}{ (6,2)}& \color{red}{(6,3)}&\color{red}{(6,4)}&\color{red}{(6,5)}&\color{red}{(6,6)}\\
\end{array}$
In this case all the red cases are impossible because the sum is more than $3$.  So in this case of the $3$ possible cases (black or blue).  $1$ of them is doubles (blue).
So the probability is $\frac 1{3}$.
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In your post the first image is:
What is the probability that BOTH faces are the Same and sum is at most 3. $P(A\&B)$
The second image is:
What is the probability that the Sum is at most three GIVEN that faces are the same.$P(B|A)$.
They are trying to show that $P(A|B)$ is the same thing as $\frac {P(A\&B)}{P(B|A)}$.
