Visualising independence of events Let's say we have a fair five-sided die. The sides of the die are numbered from 1 to 5. Each die roll is independent and all faces are equally likely. We roll twice.
Event A = the total of two rolls is 10
Event B = at least one roll resulted in 5
I get that these are clearly dependent. The $P(B\mid A) = 1$ because if you get two rolls = 10, they had to be 5 and 5, so clearly B occurs.
But how would I visualize this on a Venn diagram? Like I'm not sure how the P(B|A) = mA intersect B/P(A) and it equals 1.
To be clear, I know why the answer is so intuitively/logically but not sure how it would look visually (like Venn diagram) and mathematically.  How do we get a 1?
 A: Independence of events is not straightforward to intuit from Venn diagrams (unlike mutual exclusivity, which is observed by inspecting their intersection).
For example:

$$ \begin{array}{r}  \begin{array}{c|c|c}
  \style{font-family:inherit}{}  & \style{font-family:inherit}{U_1}  & \style{font-family:inherit}{U_2}  & \style{font-family:inherit}{U_3}
\\\hline
  \style{font-family:inherit}{P(X\cap Y)}  & 0  & \frac14  & \frac14
\\[0pt]\hline
  \style{font-family:inherit}{P(X)P(Y)}  & \frac14\times\frac12=\frac18  & \frac14\times\frac34=\frac38  & \frac12\times\frac12=\frac14
\\[0pt]\hline
  \style{font-family:inherit}{\therefore X\text{ and }Y\text{ are}\ldots}  & \textbf{dependent}  & \textbf{dependent}  & \textbf{independent}
 \end{array}\hskip-5.5pt  \end{array} $$
[Universe $U_1$ above is also an example of the fact that for events with nonzero probabilities, $\big(\text{mutual exclusivity}\implies\text{dependence}\big)$.]
Two more examples, but involving $3$ events: in each case, events $A,B$ and $C$ are pairwise independent yet are not (mutually) independent $\big($since $P(A \cap B\cap C) \neq P(A)P(B)P(C)\,\big):$



In the above universe, $a,b,c$ and $d$ denote probabilities associated with events $X$ and $Y.$ \begin{align}&\text{events }X \text{ and } Y \text{ are }\textbf{independent}
\\\iff &P(X\cap Y)=P(X)P(Y)
\\\iff &\frac{c}{a+b+c+d}=\frac{b+c}{a+b+c+d}\times\frac{c+d}{a+b+c+d}
\\\iff &ac=bd.\end{align}
In particular, for the OP's given scenario, since $\left(\frac{16}{25}\right)\left(\frac{1}{25}\right)\neq\left(\frac{0}{25}\right)\left(\frac{8}{25}\right),$ events $A$ and $B$ are dependent.


When the probability experiment has just $2$ trials, a table like this is a good way to understand/visualise conditional probability as working in a reduced sample space:
$$ \begin{array}{r}  \begin{array}{c|c|c}
  \style{font-family:inherit}{\text{time of complaint}\bigg\\ \text{reason for complaint}}  & \style{font-family:inherit}{\textbf E\text{lectrical}}  & \style{font-family:inherit}{\textbf M\text{echanical}}  & \style{font-family:inherit}{\textbf L\text{ooks}}
\\\hline
  \style{font-family:inherit}{\textbf D\text{uring guarantee period}}  & 18\%  & 13\%  & 32\%
\\[0pt]\hline
  \style{font-family:inherit}{\textbf A\text{fter guarantee period}}  & 12\%  & 22\%  & 3\%
 \end{array}\hskip-5.5pt  \end{array} $$
The calculation (notice that the figure ‘$32$’ was obtained from the intersection of column $L$ and row $D$) $$P(L|D)=\frac{P(L\cap D)}{P(D)}=\frac{32}{18+13+32}=51\%\neq32\%+3\%=P(L)$$ shows that $L$ and $D$ are dependent events.
A: As you noted, $A$ is a subset of $B$, which implies $A \cap B = A$. So, $$P(B \mid A) = P(B \cap A) / P(A) = P(A) / P(A) = 1.$$
A: Don't use Venn.  Consider instead this table:

If you're in the red, what is the probability you're in the green (overlapping red) region?
Clear?
