3
$\begingroup$

The addition law for non mutually exclusive events is given in my textbook as

$P(A \space \text{or} \space B)=P(A \space\cup \space B)= P(A)+P(B)-P(A \space \cap \space B)$

I understand the logic behind this and would be fine if it were written as $P(A \space \text{or} \space B)=P(A)+P(B)-P(A \space \cap \space B)$ however I am fairly certain that in this case it is incorrect to say that $P(A \space \text{or} \space B)=P(A \space\cup \space B)$

Here is a diagram to further elaborate A union B

From the diagram would it not be true that $P(A \space \cup \space B)=P(A \space \text{or} \space B) + P(A \space \text{and} \space B)$?

Is there a mistake in my textbook or am I missing something crucial?

$\endgroup$

3 Answers 3

3
$\begingroup$

In this context “A or B” means “A, or B, or both”. So it is correct to say that $P(A \text{ or } B) = P(A \cup B)$. There’s no mistake in your textbook.

$\endgroup$
4
  • $\begingroup$ If that is the case then why have they subtracted the $P(A \space \cap B)$ term , would $P(A \space \cap B)$ not qualify as both A,B? $\endgroup$ Commented Apr 26, 2021 at 10:59
  • 1
    $\begingroup$ Because if you just add P(A) and P(B), you’ll be counting the intersection area twice. You have to subtract one of the two copies. $\endgroup$
    – bubba
    Commented Apr 26, 2021 at 11:52
  • 2
    $\begingroup$ Forget about probabilities for a minute — just think about areas. $\endgroup$
    – bubba
    Commented Apr 26, 2021 at 11:53
  • 2
    $\begingroup$ Area of overall figure-eight = area of A + area of B - area of overlap. $\endgroup$
    – bubba
    Commented Apr 26, 2021 at 11:54
2
$\begingroup$

To be more precise...

$$P(x \in A ~\text{or}~ x \in B) = P(x \in (A \cup B)).$$

Bonus:

$$P(x \in A ~\text{and}~ x \in B) = P(x \in (A \cap B)).$$

$\endgroup$
1
$\begingroup$

It seems by the definition that when they talk about $P(A\text{ or }B)$ they actually mean $P(\text{at least }A\text{ or }B)$, that is $P(A,B\text{ or }A\text{ and }B$)

$\endgroup$
1
  • 1
    $\begingroup$ Yes, "or" is used in the inclusive sense. Exclusive or can be written "xor". So A xor B means A or B but not both. $\endgroup$
    – GEdgar
    Commented Apr 26, 2021 at 10:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .