Question

I know that we can say P(A or B) = P(A) + P(B) - P(A and B)

But can we say P(A and B) = P(A) + P(B) - P(A or B)

?


also can we say that P(A' and B') = 1 - P(A and B)

,

P(A' or B') = 1 - P(A or B)

Answer 1

no because "or" means multiply; "and" means add. Adding and Multiplying are not interchangeable

Answer 2

I saw it in a paper

Answer 3

okay...? what does that have to do with anything

Answer 4

idk .. just wonderin

Answer 5

oh lol. Well there is your answer

Answer 6

also can we say that P(A' and B') = 1 - P(A and B)

,

P(A' or B') = 1 - P(A or B)

Answer 7

I dont think so... primes (the little ') are very different

Answer 8

' doesnt indicate prime , they are the complements

Answer 9

well in calculus, the ' indicates prime and that was my first thought

Answer 10

but you could simplify
( A and B ) ' = A' or B'

Answer 11

that is true from set theory

Answer 12

$$ \Large {P(A \cup B) = P(A) + P(B) - P(A\&B)
\\~\\ \Large P(A \& B ) = P(A) \cdot P(B|A)
\\~\\
P(A ' ) = 1 - P(A)
\\ \therefore \\
\Large P[(A \& B)' ] = 1 - P( A\&B)
\\ \Large P[(A \cup B)' ] = 1 - P( A\cup B)
\\ \therefore \\
\large P[(A \& B)' ] = P[ A' \cup B' ] = P(A' ) + P(B') - P( A' \& B' )
\\ \large P[(A \cup B)' ] =P[ A' \& B' ] = P(A' )\cdot P(B' | A ' )

}
$$

Answer 13

some authors use \( \Large \cap \) for `and`

Answer 14

@mastermindkakashi
'or' implies add in the formula.
'and' means multiply

Answer 15

ah okay thanks for clarifying @perl
though you still can't change them up

Answer 16

right, they have separate formulas.
substitution has to be done consistently

Answer 17

Actually the answer is very simple

Start here, each of these things can be thought of as simply representing numbers and what you can do with numbers works perfectly fine with them, so we can do regular algebra on it.

P(A or B) = P(A) + P(B) - P(A and B)

So we can just add P(A and B) to both sides to get

P(A or B) +P(A and B) = P(A) + P(B)

Now subtract P(A or B) from both sides and we get the equation you were looking for.

P(A and B) = P(A) + P(B) - P(A or B)

I think the best way to visualize this particular instance if you really want to is with venn diagrams.

Answer 18

P(A' or B') = 1 - P(A or B) ?
P(A' and B') = 1- P(A and B) ?

@Kainui @perl

Answer 19

no christos, that is false (generally)

Answer 20

$$
\Large A' ~or~ B' \neq (A ~or~ B ) ' \\
\Large \color{blue}{\text {But...}\\}
\\ \Large A' ~or~ B' = (A ~and~ B ) '
\\ \Large A' ~and~ B' = (A ~or~ B ) '
$$
Therefore:
$$
\Large P(A' ~or~ B') =P( (A ~and~ B ) ') = 1 - P( A ~and~ B )
\\~\\
\Large P(A' ~and~ B') =P( (A ~or~ B ) ') = 1 - P( A ~or~ B )
$$

Answer 21

thank you