Question

P(E) =.30 P(F)= .09 P(E u F)= .36
find the conditional probability P(EIF)

Answer 1

I assume P(E|F) means the probability of E or F or both.

Answer 2

both

Answer 3

would it be a venn diagram?

Answer 4

No, we can do it this way:

\[P(E \cup F) = P(E) \cdot P(E|F)\]\[\implies P(E|F) = \frac{P(E\cup F)}{P(E)}\]

Answer 5

so just .36/.30?

Answer 6

?

Answer 7

Yep

Answer 8

that was incorrect...

Answer 9

Ack! sorry

Answer 10

first i need to find P(E n F) and divide that by P(F)

Answer 11

positive?

Answer 12

No. Lemme check something

Answer 13

No, that's wrong.

Answer 14

Ok, I have it.

\[P(F) + P(E) - P(E\cup F) = P(E\cap F)\]

Answer 15

That's from the venn diagram

Answer 16

polpak i think you wrote your first equation wrong

Answer 17

I did.

Answer 18

And a few other ones too. It's not my day for probability.

Answer 19

It shoulda been:
\[P(E\cap F) = P(F) \cdot P(E|F)\]

Answer 20

yep
i thought \cup was intersection
what is it for intersection

Answer 21

\cap

Answer 22

oh

Answer 23

\cup is union, (it looks like a u)

Answer 24

\[P(E|F)=\frac{P(E \cap F)}{P(F)}\]

Answer 25

Yep, then we just had to find \(P(E\cap F)\)

Answer 26

And I think I had the right reasoning for that.

Answer 27

\[P(E \cup F)=P(E)+P(F)-P(E \cap F) \]
if we solve this for P( E n F) we will get what polpak has above

Answer 28

Ah, good. I reasoned it a different direction, but arrived at the same place =)

Answer 29

\[P(E \cap F)=P(E)+P(F)-P(E \cup F)\]

Answer 30

so we need to do
\[P(E|F)=\frac{P(E)+P(F)-P(E \cup F)}{P(F)}\]