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Homework Help: Independent Probability

  1. Mar 5, 2008 #1
    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. jcsd
  3. Mar 5, 2008 #2

    tiny-tim

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    Hi pace! :smile:

    (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? :confused:
     
  4. Mar 5, 2008 #3
    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
  5. Mar 5, 2008 #4

    tiny-tim

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    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? :redface:

    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? :smile:
     
  6. Mar 5, 2008 #5
    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?
     
  7. Mar 5, 2008 #6
    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
  8. Mar 5, 2008 #7

    tiny-tim

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    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? :smile:
     
  9. Mar 5, 2008 #8
    Ok, no I really meant multiplied. I'll be back. Thanks a lot for help. Yeah, thinking like that helps me sometimes.
     
  10. Mar 6, 2008 #9
    7/8 + 8/8 ? o_O 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.
     
  11. Mar 6, 2008 #10

    tiny-tim

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    fractions!

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

    7/8 + 8/8 = 15/8. :frown:

    Try again! :smile:
     
  12. Mar 6, 2008 #11
    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..
     
  13. Mar 6, 2008 #12
    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
  14. Mar 6, 2008 #13

    tiny-tim

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    … yes … !! :smile:

    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? :smile:
     
  15. Mar 6, 2008 #14
    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 o_O?! 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 ) o_O
     
    Last edited: Mar 6, 2008
  16. Mar 6, 2008 #15
    No, where's the RR?
     
  17. Mar 6, 2008 #16
    What's 7/8 ? (checking my books)
     
  18. Mar 6, 2008 #17

    tiny-tim

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    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)? :smile:
     
  19. Mar 6, 2008 #18
    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
  20. Mar 6, 2008 #19

    tiny-tim

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    No no no …

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

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

    Try again … begin a sentence "(1/8)x(8/8) is the probability of … " :smile:
     
  21. Mar 6, 2008 #20
    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
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