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Probability problem involving binomial expnasions

  1. Oct 16, 2008 #1
    1. The problem statement, all variables and given/known data

    a) Expand ([2/3]+[1/3])^4
    b) Four chocolates are randomly taken (with replacement) from a box containing strawberry creams and almond centres in the ratio 2 : 1. What is the probability of getting:
    i) all strawberry creams
    ii) two of each type
    iii) at least 2 strawberry creams?

    (I can't use the things at the top of the form so I'll write fractions as (x/y) and powers as x^y)


    2. Relevant equations


    ...well, this is how you do the general binomial expansion:
    Tr+1 = (n,r) a^(n-r) b^r

    n is the number of terms
    r is the term you are at
    the r+1 is supposed to be subscript
    the n & r in (n,r) (that is the coefficient) are supposed to be one above the other

    3. The attempt at a solution

    a) is easy: ([2/3]+[1/3])^4 = (2/3)^4 + 4(2/3)^3(1/3) + 6(2/3)^2(1/3)^2 + 4(2/3)(1/3)^3 + (1/3)^4

    b) is also mostly easy:
    i) (2/3)^4 = (16/81)
    ii) 6(2/3)^2(1/3)^2 = (8/27)
    iii) this one I can't do. The book says the answer is (8/9) but I don't know how they got it. I got (392/729). This is how i got it. The question asks for possibilites of 2 or more of the 2/3 chocolates, so:
    (2/3)^4 + 4(2/3)^3(1/3) + 6(2/3)^2(1/3)^2 =
    = (16/81) + 4(8/27)(1/27) + (8/27)
    = (16/81) + (8/27)(4/27) + (24/81)
    = (144/729) + (32/729) + (216/729)
    = (392/729)
    I don't know what I do wrong, but I have a feeling it's in the probability part. There's probably something wrong with how I'm combining the probilities of 2 strawberry, 3 strawbeery & 4 strawberry. Can anyone lend me a hand please? Is there anything more that I need to add?

    thanks in advance
    -sinisterstuf
     
  2. jcsd
  3. Oct 16, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi sinisterstuf ! Welcome to PF! :smile:
    erm … where did those 27s come from?? :confused:

    all the fractions should be over 81, shouldn't they? :smile:
     
  4. Oct 16, 2008 #3

    HallsofIvy

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    Okay. That is 6(4/9)(1/9) and you have canceled a "3" in both numerator and denominator, right?

    (16/81)+ 4(8/27)(1/3)+ 6(4/9)(1/9) = 16/81+ 32/81+ 24/81= 72/81

     
  5. Oct 16, 2008 #4
    Wow thanks. And thanks for the welcome :)

    and HallsofIvy is right. It's to get the answer in simplest form.

    Ok, so you get the correct answer but there's a part i don't understand. How do you get the 32/81?! I understand that 4(8/27) = 32/81, right? But where does the 1/3 go then? The one that's next to the 4(8/27) to be multiplied with it. Am I just tired or am I just nor getting this?
     
  6. Oct 17, 2008 #5

    tiny-tim

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    … it was lost, but now it's found …

    Hi sinisterstuf! :smile:

    Yes, you're very tired, or you wouldn't have made all those mistakes at the start.

    4(8/27) = 32/27, not 32/81.

    The 1/3 wandered away when you dropped off for a few minutes. :wink:

    … you take your eye off them for a moment :rolleyes:

    Useful tip: in problems like this, the denominators will always be the same … so check each one as you go, and if it's not (in this case) 81, then you know you've made a mistake.

    When you're herding fractions, remember …

    fractions of a feather flock together! :smile:

    Another useful tip: Get some sleep!! :zzz:
     
  7. Oct 17, 2008 #6
    Hahaha, oh my goodness that's embarrasing, I feel a bit... dumb. But I was so tired, it's my excuse. :P

    Thanks for your help!
    I'll keep your advice in mind ;)

    thanks
    -sinisterstuf
     
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