If $P(A) = 0.4$, $P(B) = 0.7$, and $P(A \cap B) = 0.2$, what is the value of $P(A \cup B)$
(I'm lost pls help)

Question

If $P(A) = 0.4$, $P(B) = 0.7$, and $P(A \cap B) = 0.2$, what is the value of $P(A \cup B)$
(I'm lost pls help)

notmeta · Answer

I know the answer of the question but I need to know how to solve it.

Breathless · Answer

hold on

Breathless · Answer

addition rule of probability as so you would take A and B and do A+B as of which it doesn't matter what numbers u put in those spots all we care about is subtracting the other number, so if 0.7+0.4 is what we have then we will minus it by 0.2 so in turn we have 0.7+0.4-0.2 which equals 0.9

Breathless · Answer

if not am stumped

Breathless · Answer

here's what Google has to say:

To find the value of $P(A \cup B)$, you use the Addition Rule for probability.

The Formula:$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

The Calculation:Plug in your values: $0.4 + 0.7 - 0.2$
Add the individual probabilities: $0.4 + 0.7 = 1.1$
Subtract the intersection: $1.1 - 0.2 = 0.9$The value of $P(A \cup B)$ is 0.9.

Why do we subtract $P(A \cap B)$?

Think of it like two overlapping circles in a Venn diagram:When you add $P(A)$ and $P(B)$ together, you are counting the area where they overlap (the intersection) twice.To get the correct total area (the union), you have to subtract that overlap once so it's only counted once.