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## Homework Statement

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)

## Homework 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

## 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