# Probability problem involving binomial expnasions

## 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?

-sinisterstuf

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tiny-tim
Homework Helper
Welcome to PF!

Hi sinisterstuf ! Welcome to PF! (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)
erm … where did those 27s come from?? all the fractions should be over 81, shouldn't they? HallsofIvy
Homework Helper

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

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

= (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?

-sinisterstuf

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?

tiny-tim
Homework Helper
… it was lost, but now it's found …

… 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?
Hi sinisterstuf! 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. … you take your eye off them for a moment 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! Another useful tip: Get some sleep!! :zzz:

Hahaha, oh my goodness that's embarrasing, I feel a bit... dumb. But I was so tired, it's my excuse. :P