MHB How Do You Calculate P(A∩B) in Probability Theory?

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To calculate P(A∩B) in probability theory, the formula P(A∪B) = P(A) + P(B) - P(A∩B) is used. By rearranging this formula, P(A∩B) can be isolated as P(A∩B) = P(A) + P(B) - P(A∪B). In the given example, substituting the values leads to P(A∩B) = 1 - 0.6 - 0.8, resulting in a negative probability, indicating a mistake in the values used. Additionally, the discussion touches on De Morgan's laws, which relate the complements of unions and intersections, helping to clarify the notation and calculations involved. Understanding these concepts is crucial for accurate probability calculations.
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this is an image of the problem as given, mostly to avoid typo's, but (b) seems to contain a notation that I don't recognize.

well for $$(a)$$ find $$p(A\cap B)$$

from the counting Formula:
$$n(A\cup B)=n(A)+n(B)-n(A\cap B)$$

replacing $n$ for $p$ and isolating $$p(A\cap B)$$
$$p(A\cup B)-p(A)-p(B)=p(A\cap B)$$
so..
$1-0.6-0.8=-0.4=p(A\cap B)$

(b) ?
 
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Hi karush,

For (a) you have the right idea but mixed up your signs a bit.

$P[A \cup B]=P[A]+P-P[A \cap B]$. We want to solve for $P[A \cap B]$.

$P[A \cap B]=P[A]+P-P[A \cup B]$.

Can you solve it from there?

(b) I'm going to guess that the "C" means "compliment". Normally we write this as $P[A' \cup B']$ or $P[A^c \cup B^c]$. There are some special and useful rules, called De Morgan's laws, that allow us to manipulate expressions like this. Have you seen these before?
 
$$P[A \cap B]=P[A]+P-P[A \cup B]$$

$$0.4=0.6+0.8-1$$

i am am looking at De Morgan's laws it new to me...

I did see this

https://www.physicsforums.com/attachments/1136
 
Exactly! :)

Notice that $P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]$ and you'll be done.
 
Jameson said:
Exactly! :)

Notice that $P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]$ and you'll be done.

$$P[A' \cup B']=P[A \cap B]'=1-P[A \cap B]$$
$$

0.6=1-0.4$$

really that it!
 
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