# Independent Probability

1. Mar 5, 2008

### pace

uh, I mean probability with independent outcomes, heh.

1. The problem statement, all variables and given/known data

A wheel I'm spinning twice has 8 different colours(among them the colour red). It will then give two different outcomes. Q: What is the probability that the outcome will give the colour red at least once.

3. The attempt at a solution

((1/8)x(7/8))x2/3

or

((1/8)x(7/8))x((1/8)(1/8))

Neither gives the result 0.234, which is the answer. 0_o

Last edited: Mar 5, 2008
2. Mar 5, 2008

### tiny-tim

Hi pace!

(1/8)x(1/8) is the probability for both spins red.

(1/8)x(7/8) is the probability for first-spin-not-red & second-spin-red.

But where does your 2/3 come from?

3. Mar 5, 2008

### pace

Hi there :)

Oh sorry, the last one I meant ((1/8)x(7/8))x((1/8)(1/8))

I'm thinking the 2/3 is the "red at least once".......

Last edited: Mar 5, 2008
4. Mar 5, 2008

### tiny-tim

umm … ((1/8)x(7/8))x((1/8)(1/8)) is the probability for first-spin-not-red & second-spin-red & thrid-spin-red & fourth-spin-red … but there's only two spins!

You're really confused, aren't you?

Let's see … make the question as simple as possible … let's write R for red, and N for not-red.

Then there's four possiblities: RR, RN, NR, and NN.

You're only interested in the total of the first three (ie, not NN).

So how do you count them?

5. Mar 5, 2008

### pace

Ah, that gives four spins..! Yep, it's a mess in here :) I wrote down two answers that seemed the closest to what I felt would give the answer, eheh.

Um, ah yes, that gives 3/4, yes?

6. Mar 5, 2008

### pace

hm, maybe I see, I start at of at the wrong place? I started thinking off at P(a)xP(b).

But...um.. so.. RR((1/8)x(1/8)) x RN(1/8)x(7/8) x NR(7/8)x(1/8) doesn't work ???

But then I miss Probable/Possible(NN being here)

(Work. I'm thinking I should maybe do more of the easier ones first)

Last edited: Mar 5, 2008
7. Mar 5, 2008

### tiny-tim

pace, why do you keep writing "x"?

(oh, you did that in your ((1/8)x(7/8))x((1/8)(1/8)) also, I think)

It's "+".

(btw, your answer 3/4 would have been right if it was a coin, so I think you've got the principle.

But you don't seem able to write it clearly in mathematics.)

Hint: 0.234 = 15/64 - does that help?

8. Mar 5, 2008

### pace

Ok, no I really meant multiplied. I'll be back. Thanks a lot for help. Yeah, thinking like that helps me sometimes.

9. Mar 6, 2008

### pace

7/8 + 8/8 ? The.. 7/8 would stand for no red, the 8/8 for all red, but that wouldn't make sense.. 7/8 + 4/8 + 4/8.. no.

Um.. arg, I don't get it.

10. Mar 6, 2008

### tiny-tim

fractions!

Oh pace, you're no good at fractions, are you?

7/8 + 8/8 = 15/8.

Try again!

11. Mar 6, 2008

### pace

hehe sorry. I don't know where to begin maybe. Well I have a B from 1mx actually, but it is a bit too long ago. I'm having a hard time only with probablity .

um, (1/8)x(8/8)+(7/8)(1/8) ?!.. no..

12. Mar 6, 2008

### pace

That gives two results. I'm leaving out the 3/4(Probable/Possible(sp?!)) of course, but I don't seem to make that add up: Like (1/8)x(8/8)+(7/8)(1/8) (a third spin here? But that would make over 15). And then add four divisions at the bottom(Possible results), but that would makes a lot of counting, and it doesn't seem to me that that would be the way.

Last edited: Mar 6, 2008
13. Mar 6, 2008

### tiny-tim

… yes … !!

ok, work backwards - you know it's (1/8)x(8/8)+(7/8)(1/8).

So, what is (1/8)x(8/8) the probability of?

And what is (7/8)x(1/8) the probability of?

14. Mar 6, 2008

### pace

lol. I don't get the 8/8. .. ah it's 'no red' of course... right? *hits my own head* Why didn't I think that way... I got to think math to language/pictures ?! I'm afraid of probability. Cause I'm much better at maths than language and probablity seems to me to be more language oriented(?) blah blah blah.

(1/8)x(8/8) is NRxNN + (7/8)x(1/8) RRxRN.

But I was sure I had to bring in the Probable/Possible in there somewhere.. With a whole ( math ) / ( math )

Last edited: Mar 6, 2008
15. Mar 6, 2008

### pace

No, where's the RR?

16. Mar 6, 2008

### pace

What's 7/8 ? (checking my books)

17. Mar 6, 2008

### tiny-tim

No!

N is no-red. R is red.

So NR is first-not-red + second-red.

And NRxNN would be first-not-red + second-red + third-not-red + fourth-not-red, if you had 4 spins (and you wouldn't need to write the "x").

So … try again … what is (1/8)x(8/8)? … and then … what is (7/8)x(1/8)?

18. Mar 6, 2008

### pace

checked.

"N is no-red. R is red.", yes, this I get... something like it.

You mean a x between second-red and third-not-red right?

at least one red x no red + (but shouldn't it stand 2/8 here, that would give probabllity for at least two reds) x at least one red. gaaawd, I'm so confused.

Last edited: Mar 6, 2008
19. Mar 6, 2008

### tiny-tim

No no no …

That doesn't even make any sense, does it?

(if there's at least one, how can there be none? )

Try again … begin a sentence "(1/8)x(8/8) is the probability of … "

20. Mar 6, 2008

### pace

RNxNR:(R(at least one red))(N(no red))+(N(No red))(R(at least one red)) . But where's the RR and the other NR?

... (R(1/8))(N(8/8)+(N(7/8)x(R(1/8)).... No..

or R(probablility of at least one red)N(probability of no red) + NR huh.

... Thinking :)

Last edited: Mar 6, 2008