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?

Solution 1:

Independence of events is not straightforward to intuit from Venn diagrams (unlike mutual exclusivity, which is observed by inspecting their intersection).

For example: enter image description here

$$ \begin{array}{r} \begin{array}{c|c|c} \style{font-family:inherit}{} & \style{font-family:inherit}{U_1} & \style{font-family:inherit}{U_2} & \style{font-family:inherit}{U_3} \\\hline \style{font-family:inherit}{P(X\cap Y)} & 0 & \frac14 & \frac14 \\[0pt]\hline \style{font-family:inherit}{P(X)P(Y)} & \frac14\times\frac12=\frac18 & \frac14\times\frac34=\frac38 & \frac12\times\frac12=\frac14 \\[0pt]\hline \style{font-family:inherit}{\therefore X\text{ and }Y\text{ are}\ldots} & \textbf{dependent} & \textbf{dependent} & \textbf{independent} \end{array}\hskip-5.5pt \end{array} $$ [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 (mutually) dependent $\big($since $P(A \cap B\cap C) \neq P(A)P(B)P(C)\,\big):$ enter image description here

enter image description here

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 $(16)(1)\neq(0)(8),$ events $A$ and $B$ are dependent.

enter image description here

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.