Question

how to prove P(EUF) = P(E) + P(F) –P(EF) by P(x) i mean Probability of x

Answer 1

Okay, what can we assume is true? Are we going back to axioms here?

Answer 2

Yes .We have to prove it using the Axioms here

Answer 3

Can you tell me the axioms you think will help us?

Answer 4

0 ≤ P(E) ≤1
P(S) =1
and axioms for mutually excusive event

Answer 5

*exclusive

Answer 6

S is universal set?

Answer 7

S is sample space

Answer 8

Okay, I think we'll need to pull out set theory.

Answer 9

\[
\Pr(E) = \frac{\|E\|}{\|S\|}
\]

Answer 10

Can we say: \[
\|E\cup F\| = \|E\|+\|F\| -\|E\cap F\|
\]

Answer 11

i have no idea abut this

Answer 12

Do you know what the cardinality of a set is?

Answer 13

The number of elements in a set.

Answer 14

ok

Answer 15

Do we need to prove that formula I gave?

Answer 16

i need to prove this P(EUF) = P(E) + P(F) –P(EF)

Answer 17

using axioms

Answer 18

You also need to use the definitions as well.

Answer 19

How about this: \[\begin{split}{}
\Pr(E\cup F)
&= \frac{|E\cup F|}{|S|} \\
&= \frac{|E|+|F|-|E\cap F|}{|S|} \\
&= \frac{|E|}{|S|} +\frac{|F|}{|S|} -\frac{|E\cap F|}{|S|} \\
&=\Pr(E)+\Pr(F)-\Pr(E\cap F)
\end{split}
\]

Answer 20

This is a definition. We can always use definitions:\[
\Pr(A) = \frac{|A|}{|S|}
\]

Answer 21

The only thing that might need more rigor is \[
|E\cup F|=|E|+|F|-|E\cap F|
\]

Answer 22

How about this?
\[A\cup B = A\cup (A'\cap B)\]So, \[P(A\cup B) = P(A)+ P(A'\cap B)\]
Note: A and A'∩B are disjoint.
Also,
\[B = (A\cap B) \cup (A'\cap B)\]
So,
\[P(B) = P(A\cap B) + P(A'\cap B)\]
Note: (A∩B) and (A′∩B) are also disjoint.

Combine the two results, it should give you what you need to prove.

Answer 23

use the relations in sets