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