Background/Context: I tried to solve the question

For a certain probability experiment, the probability that event $A$ will occur is $\frac 12$ and the probability that event $B$ will occur is $\frac 13$. Which of the following values could be the probability that the event $A \bigcup B$ will occur? (select all that apply) $a) \frac 13$ $b) \frac 12$ $c) \frac 34$

Since the question gives no information about the dependency I assume both extreme cases:

[Independent] $p(A \bigcup B) = p(A) + p(B) = \frac 12 + \frac 13 = \frac 56$
[Dependent] $p(A \bigcup B) = p(A) + p(B)-p(A \bigcap B) = \frac 56 - p(A \bigcap B)$

Here is where I went wrong, assuming $p(A \bigcap B)$ can take on any value between $0$ and $\frac 56$, making all the choices possible. An explanation states: "The lower limit will occur when the events are dependent and therefore p(A and B) will equal the smallest of the two probabilities. Thus, the lower limit will be whichever probability is greater between p(A) and p(B)"

Midway through the question I thought $p(A \bigcap B)$ can be any value because I just imagined the intersection between two sets on a Venn diagram getting bigger till they completely overlap. I now realise that's completely wrong since a Venn diagram illustrates whether the events are exclusive or not.

But I can't visualize or come up with an example that demonstrates the explanation. How can I see that when events are completely dependent the result will be the larger probability of the two events? Can someone explain?

  • 2
    $\begingroup$ $\frac56$ is the highest possible value, but they are not then independent; instead they are mutually disjoint. Meanwhile $P(A \cap B)$ cannot be larger than either $P(A)$ or $P(B)$ since it is based on an intersection $\endgroup$
    – Henry
    Commented May 12, 2019 at 18:46
  • $\begingroup$ $P(A \cup B) = P(A) +P(B)$ holds for disjoint events, not independent events. Furthermore, $A$ and $B$ can be independent and still have $A \cap B \neq \emptyset$. Therefore the second case should be called not disjoint rather than dependent. $\endgroup$
    – kccu
    Commented May 12, 2019 at 18:50

1 Answer 1


As Henry and kccu pointed out, disjoint events have: $$ A \cap B = \{ \} \implies P(A \cap B) = 0 $$

Therefore, if $A$ and $B$ are disjoint, then their union will have the maximum possible probability $$ P(A \cup B) = P(A) + P(B) $$

On the other hand, if $A$ and $B$ overlap as much as possible, then their interception will correspond to the smallest one $$ P(A \cap B) = \min \Big( P(A), P(B) \Big) $$ and their union will correspond to the largest one $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) = \max \Big( P(A), P(B) \Big) $$

Can you now find the interval of possible values for $P(A \cup B)$?

  • $\begingroup$ Wow I think I get it. I had trouble seeing or realising 𝑃(𝐴∩𝐵)=min(𝑃(𝐴),𝑃(𝐵)). But essentially, if I am looking at it in terms of a Venn diagram it would be a smaller circle engulfed by a larger circle, yes? So the interception is just essentially the smaller circle (probability). $\endgroup$
    – Bn.F76
    Commented May 12, 2019 at 19:06
  • $\begingroup$ Yes, that's it the smaller set is inside the larger set. If you think it's appropriate, you may accept my answer (there is a green button next to it). Or you may way for a day or so to see if more helpful answers appear. $\endgroup$ Commented May 12, 2019 at 19:08

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