1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
  2. jcsd
  3. Oct 16, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    … 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 ;)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook