Question

What does P(E|F) mean?

Answer 1

Probability of E given F.

Answer 2

i know p(e) is prob of e, what does the line represent

Answer 3

Would u be able to give an expample?

Answer 4

that notation denotes "conditional probability."
Given that event F has already happened, what is the probability that Event E will happen?

This topic, while not really difficult, does take time to learn. Have you read your textbook or online learning materials about "conditional probability?" I'm sure there'd be examples there. Others and I could give you a host of examples, but I'd much prefer you do some background reading on this topic if you haven't already.

Answer 5

I love examples. :3
Ever heard of that casino game "roulette" ?

Answer 6

yes!

Answer 7

Of course, probability of the outcome of a dice thrown given it is an even number is 1/3. If this was not given the probability would have been 1/6. Here E is final event and F is initial event. The line is read as "given that" another event has occurred. .

Answer 8

Does that make sense?

Answer 9

Too bad none of it is universal @askme12345
Let these events be represented as follows:
O = odd
E = even
R = red
B = black

Answer 10

Understood @askme12345 ?

Answer 11

oh okay! thanks that simplifies it. can anyone explain, how i know when to use, the disjoint formula as opposed to the multipication or addition rule? How would i know when to use what

Answer 12

P(E | F) = "the probability that event E will occur, given that (if we know) that event F has already occurred." Again, if you want to develop any depth in your undrstanding of conditional probability, please look for definitions and examples in your course reading materials.

Or, you could start by telling US what YOU already know about conditional probability. Then someone might fill you in on what further you need to know.

Answer 13

Oh, never mind examples then :(
Back to the world of equations lol

Answer 14

In case the events are independent ie disjoint, you can use the multiplication theorem.

Answer 15

@terenzreignz i would still like to hear your example lol! just trying to clarify a lot of things on this topic

Answer 16

Really? Because dice are simpler (and more universal)
But I could ask you a few certain things if we use a specific wheel for reference.
This might be tedious, but what the heck XD

http://en.wikipedia.org/wiki/File:European_roulette_wheel.svg

Look closely... 37 possible outcomes, and what not, what's the probability of the ball landing on black?
P(B) = ?

Answer 17

18/36

Answer 18

18/37 *
I'd love to say close enough, but your instructor probably abides by "close doesn't count" instead :P

But anyway, that's all good and well, but what about the probability that the ball lands on an EVEN number? (0 counted as even)
P(E) = ?

Answer 19

19/37

Answer 20

That is correct. So far we haven't seen that vertical line that irks you yet (lol)
So let's use it.

Now, here's a more 'fun' example.

Given that the ball has landed on black, what is the probability that it lands on an even number?

In symbols, we write:
P(E | B) = ?

(Probability of the ball landing on an even number, given that it landed on black)

Answer 21

11/37
Question! I figured that out by counting, but how would i do it givent he prob of the even and the black, lets say i didnt have the pic or it was a dif example

Answer 22

It's 10/37.
I think.
0 is not black.

Answer 23

Oh, say you were just given that P(B) = 18/37
and P(E) = 19/37, right?

Answer 24

yup you're right sorry .. yup!

Answer 25

Well then, those two bits of info are not sufficient to get you the exact value of P(E | B)

Answer 26

You also need the probability of the ball landing on even AND black.

In symbols, you need \(\large \mathbb P(E\cap B)\)

Answer 27

Wait a sec... it's not 10/37 either :P

So sorry.

Backtrack... you still with me?

Answer 28

@askme12345
there?

Answer 29

yup

Answer 30

It's not 10/37, because we no longer have to look at all 37 possibilities.

Remember, we KNOW that the ball landed on black, so isn't 10/37, rather
P(E | B) = 10/18 = 5/9

Since there are 18 black spaces.

Got it? ^_^

Answer 31

ooooo right okay!

Answer 32

Okay, so, ready to translate all this into maths equations?

Answer 33

hope so haha

Answer 34

Of course you are ^_^

This is an identity, you'd do well to remember it:
\[\Large \mathbb P (E|B) = \frac{\mathbb P (E\cap B)}{\mathbb P(B)}\]
Okay?

Answer 35

So, to figure out what is P(E|B), we need to know, first, the probability of B (the probability of the ball landing on black)

AND the probability of \(E \cap B\) or the probability of the ball landing on a black EVEN number.

Do we know these numbers?
You know P(B) = 18/37
What about of the ball landing on a black EVEN number?
Can you count? :)

Answer 36

10/18

Answer 37

No... just on a black even number. How many black even numbers are there?

Remember, we're getting P(E∩B) here, which means there is no |
we're simply getting the chances of the ball landing on a black even number, against all other possibilities (meaning we divide by the total 37, rather than the total black 18)

Answer 38

So, actually, \[\Large \mathbb P(E\cap B) = \frac{10}{37}\]
agreed? ^_^

Answer 39

By the way, with probabilities concerned, you can more-or-less treat the union \(\cup\) symbol as 'or' and the intersection symbol \(\cap\) as 'and'

So P(E∩B) = probability of black AND even.

^_^

So... so far so good?

Answer 40

Oh Okay just got a little confused - yup back on track l

Answer 41

Okay, so we get our answer by substituting:
\[\Large \mathbb P (E|B) = \frac{\mathbb P (E\cap B)}{\mathbb P(B)}=\frac{\frac{10}{37}}{\frac{18}{37}}= \frac{10}{18}=\frac{5}{9}\]
Which is the answer we got earlier ^_^

Answer 42

oh okay!works out

Answer 43

Okay, quick question...
Say we let G = probability that the ball lands on green.

Then what is P(O | G) = ?

That is to say, what is the probability that the ball lands on an odd number, given that it landed on green?

Answer 44

0/1? theres one green spot and it is even (0)

Answer 45

0/1 is what? ;)

Answer 46

0?

Answer 47

YES!
LOL

What that says, basically, is that if we know the ball landed on a green spot, there's no way in hell that it landed on odd, so the probability of it being odd, given green, is zero

See? You'll get the hang of this in no time, with a little practice ;)

Answer 48

Okay haha makes sense thanks! you clarified this

Answer 49

I do try. Well, I guess that about sums up this little crash course on conditional probability... and roulette.
Good luck betting XD

jk

Answer 50

This isnt over yet? :P Too many notifications lol

Answer 51

Sorry. In retrospect, OS should seriously consider that unsubscribe option :)

Answer 52

^ :) Anyway good work teren!

Answer 53

TJ!
<evil glare>

And thanks ^_^

Answer 54

Well, anyway, @askme12345
good job, I'll be signing off now, so just hang in there, lol ^_^

----------------------------------------------------
Terence (not Teren, d****t :P ) out ;)

Answer 55

@askme12345 : If you'll forgive me ... may I suggest, again, that you actually do some reading in advance? You could have done that in the time this conversation has taken.