Melody

You will sart learning factorisation around year 8       13 years old

Permutations come in year 10 or 11                         15-16 years old

No, neither do I  MG.

Try watching this.

If you are writing in the input box you can enter

sqrt(5)     which will give you     $${\sqrt{{\mathtt{5}}}}$$             Note that this is the same as  5^(1/2)          $${{\mathtt{5}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$

OR

sqrt(5,3)   which will give you     $${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{5}}}}$$          Note that this is the same as  5^(1/3)          $${{\mathtt{5}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

Thanks Chris

Thanks Chris

$${\mathtt{44}}{\mathtt{\,\times\,}}{\mathtt{3}} = {\mathtt{132}}$$     eggs are needed.

$${\frac{{\mathtt{132}}}{{\mathtt{23}}}} = {\mathtt{5.739\: \!130\: \!434\: \!782\: \!608\: \!7}}$$

So you will need 6 packets

$${\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{23}}{\mathtt{\,-\,}}{\mathtt{132}} = {\mathtt{6}}$$

The children will get 3 each and Mrs Rodgers will get 6 for herself.  :)

I am going to assume the cards are replaced each time.

Let x be the number of blue and 60-x be the number of red

P(at least 2 red) >=0.5

P(0 or 1 red) <=0.5

P(all blue)+P(1R and 5 blue) <=0.5

$$\\\frac{x^6}{60^6}+6C1\left[\frac{(60-x)}{60}\right]^1\left[\frac{x}{60}\right]^5\le0.5\\\\\\
\frac{x^6}{60^6}+6\left[\frac{(60-x)x^5}{60^6}\right]\le0.5\\\\\\$$

I used Wolfram|Alpha to solve this

I do not understand your formula Nauseated.  Perhaps you could explain how you derived it?  Thanks

Anyway, i have contined this theme on a new post:

here it is   

@@ End of Day Wrap   Sat 28/3/15 Sydney, Australia Time 9:55 am     ♪ ♫

Hi all,

Great answers today from Falkirks, TayJay, Civonumzuk, Alan, Heureka, CPhill, Geno3141 and BrittanyJ.

Thank you  

Interest Posts:

    FTJ means "for the juniors"