Melody

If they cross at right angles the nit will be the width of one times the width of the other one.  

Draw it on a number grid and take a look :)

What is the cross over area for these ribons?

3*2=6 units squared

We don't waste syllables here... It is just  Ya welcome.    :))

Or we might say "No worries"   :)

1-[P(4 sixes and 2 non sixes)+(5 sixes and one non six)+P(6 sixes)]

\(1-[\binom{6}{4}(\frac{1}{6})^4(\frac{5}{6})^2+\binom{6}{5}(\frac{1}{6})^5(\frac{5}{6})^1+\binom{6}{6}(\frac{1}{6})^6]\\ 1-[15(\frac{1}{6})^4(\frac{5}{6})^2+6(\frac{1}{6})^5(\frac{5}{6})^1+(\frac{1}{6})^6]\\ 1-[15(\frac{5^2}{6^6})+6(\frac{5}{6^6})+(\frac{1}{6^6})]\\ 1-\frac{1}{6^6}[15*25+30+1]\\ 1-\frac{406}{6^6}\\ \)

1-406/6^6 = 0.9912980109739369

I didn't do it like that Chris,

I just said there are 5 books, any of them could be in the first position so the probability that any in particular one is first is 1 in 5.

Easy peazy :))

Oh there are a couple of Latex threads, they are in our sticky topics ... so everyone can find them easily  :)

Added to our LaTex thread here:

How to put different types of brackets around matrices - or no brackets at all !

http://web2.0calc.com/questions/test-2#r2

Mmm

I worked this our from here

one in five

\(\frac{1}{5}\)

If six coins are flipped, what is the probability of getting exactly two tails?

\(P(2 tails)=\binom{6}{2}0.5^6=15*\frac{1}{64}=\frac{15}{64} \)