5

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?

ryang
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3 Answers3

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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 not (mutually) independent $\big($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 $\left(\frac{16}{25}\right)\left(\frac{1}{25}\right)\neq\left(\frac{0}{25}\right)\left(\frac{8}{25}\right),$ events $A$ and $B$ are dependent.

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.

ryang
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As you noted, $A$ is a subset of $B$, which implies $A \cap B = A$. So, $$P(B \mid A) = P(B \cap A) / P(A) = P(A) / P(A) = 1.$$

angryavian
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  • I guess my question is, how does the math work here? I understand logic/intuition. But how do we know that we should take the probability of A intersection B divided by the probability of A? If A occurred for sure, its probability is 1? So what is the probability of A intersect B? Wouldn't it also be 1? How do we get that? –  Aug 16 '21 at 13:26
  • @ed84 The first equality $P(B \mid A) = P(B \cap A) / P(A)$ is the definition of $P(B \mid A)$. The right-hand side formalizes the notion of "conditioning on $A$" in terms of non-conditional probabilities. $P(A \cap B)$ is the probability of the event $A \cap B$ occurring; you don't have to worry about "$A$ occurred for sure" for that term. – angryavian Aug 16 '21 at 15:41
  • Ah. So P(B∩A) (the idea of intersection) is the equivalent of asking how event A impacts the probability of B? What is the conditioning of A? –  Aug 16 '21 at 15:54
  • @ed84 If you wanted $P(B)$, you would look at the outcomes in $B$ and compare their likelihood to the that of the set of all possible outcomes (which has probability $1$). If you condition on $A$, you basically throw away all outcomes not in $A$ and focus only on the ones in $A$. So on the numerator you only care about outcomes in $B$, but you've thrown away all outcomes not in $A$, so you are left with $B \cap A$. In the denominator you want to consider "all outcomes," but again you've thrown away anything not in $A$, so the denominator involves $A$ only. – angryavian Aug 16 '21 at 15:58
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Don't use Venn. Consider instead this table:

enter image description here

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

Clear?

David G. Stork
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  • Why is 1 to 5 all green? for a first and second toss? –  Aug 16 '21 at 13:03
  • A 5 on the first toss (i) or a 5 on the second toss (j). – David G. Stork Aug 16 '21 at 16:07
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    Oh. Isn't it all green because those are all the combinations that have a 5 in it? (i.e., 1 5, 5 1, 2 5, etc.? And 5 is red because that is the situation where the tosses are 5 5? –  Aug 16 '21 at 16:15
  • The green are where there is a 5 on EITHER the first toss OR the second toss. The red is where the is a 5 on BOTH the first toss AND the second toss. Clear? – David G. Stork Aug 16 '21 at 16:17
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    Yes. I believe that is what I said but I said it slightly more unclear? lol –  Aug 16 '21 at 16:18