Melody

Hi Imcool,

It is great that you had a go at it.  And 6 was the answer  :))

Did you look at the answer that I provided on the original question?

The answer is 2/5

Think of the cards as all being different 3 different red and 3 different blue

There are 6!  ways that these can be ordered.

now I am going to give the first person 2 cards then the second and then the 3rd.

6 choices to start

then 3 of the ones left are the other colour

now there are 4 cards the second person can get any of them

there are 2 that are a different colour for the seconds person second choice

there are 2 cards left for the 3rd peron to chose from and then just 1

so that is

6*3*4*2*2

so the prob will be  (6*3*4*2*2)/ 6!  =  2/5

A stick is broken at two points, at random.  If the length of the stick is 6, then what is the probability that all three resulting pieces are shorter than 3 units?

I'd use contour probability mapping

Let one piece be x another  y-x and the third   6-y

we know

0

y-x>0     which simplifies to    y>x

y-x<6      which simplifies to    y

and

0<6-y<6   which simplifies to  -6<-y< 0   or better still    0

Plot all those lines on a number plane.

The area in the middle conatins all the possible outcomes.  

Now you want the section inside where  

x<3

y-x<3  and

6-y<3

Find the area contained in this smaller region

Put the little area over the big area and you have your probability.

The prob that it appears 9 times or more is 1

We know it appears at least 9 times, so the prob that the 'other' roll is also a 6 is  1/6

so I think the answer is  5/6

Here is a graphical representation of what is happening.

If   \(f(x)=3x-7\)     then find    

\(f^{-1}(f^{-1}(-1))?\)

Let

\(y=3x-7\\ \frac{y+7}{3}=x\\ x=\frac{y+7}{3}\\ so\;if\;\\ f(x)=3x-7\\ then\\ f^{-1}(x)=\frac{x+7}{3}\\ f^{-1}(-1)=\frac{-1+7}{3}\\ f^{-1}(-1)=2\\ f^{-1}f^{-1}(-1)=f^{-1}(2)=9/3 = 3 \)

f(x)=3x-7

f^{-1}(f^{-1}(-1))?

\(x^2 + y^2 = 2x - 6y - 10x + 14y + 88\\ x^2 + y^2 = - 8x + 8y + 88\\ x^2 +8x+ y^2-8y = 88\\ x^2 +8x+ y^2-8y = 88\\ x^2 +8x+16+ y^2-8y +16= 88+16+16\\ (x+4)^2+(y-4)^2=120\)

This is a circle with a radius of sqrt120

So   

 \(A=\pi *120\\ A=120\pi\)

13     *5*13*___*3 = 13*17

\(\frac{17}{5*13*3}=\frac{17}{195}\)

the second one is the same except every second term is negative. You can work out if I meant the odd terms or the even ones.

\(\displaystyle (e^\frac{5\pi}{6} + e^\frac{4\pi}{3})^9\\ =\displaystyle \sum_{n=0}^9 \binom{9}{n}(e^\frac{5\pi\; n}{6}) (e^\frac{8\pi (9-n)}{6})\quad \\ = \binom{9}{0}(e^\frac{5\pi\; 0}{6}) (e^\frac{8\pi (9)}{6}) + \binom{9}{1}(e^\frac{5\pi\; 1}{6}) (e^\frac{8\pi (8)}{6}) + \binom{9}{2}(e^\frac{5\pi\; 2}{6}) (e^\frac{8\pi (7)}{6}) + \binom{9}{3}(e^\frac{5\pi\; 3}{6}) (e^\frac{8\pi (6)}{6}) + \binom{9}{4}(e^\frac{5\pi\; 4}{6}) (e^\frac{8\pi (5)}{6}) + \binom{9}{5}(e^\frac{5\pi\; 5}{6}) (e^\frac{8\pi (4)}{6}) + \binom{9}{6}(e^\frac{5\pi\; 6}{6}) (e^\frac{8\pi (3)}{6}) + \binom{9}{7}(e^\frac{5\pi\; 7}{6}) (e^\frac{8\pi (2)}{6}) + \binom{9}{8}(e^\frac{5\pi\; 8}{6}) (e^\frac{8\pi (1)}{6}) + \binom{9}{9}(e^\frac{5\pi\; 9}{6}) (e^\frac{8\pi (0)}{6})\\ = (e^\frac{72\pi }{6}) + 9(e^\frac{5\pi}{6}) (e^\frac{64\pi }{6}) + \binom{9}{2}(e^\frac{10\pi}{6}) (e^\frac{56\pi }{6}) + \binom{9}{3}(e^\frac{15\pi}{6}) (e^\frac{48\pi }{6}) + \binom{9}{4}(e^\frac{20\pi}{6}) (e^\frac{40\pi }{6}) + \binom{9}{5}(e^\frac{25\pi }{6}) (e^\frac{32\pi }{6}) + \binom{9}{6}(e^\frac{30\pi}{6}) (e^\frac{24\pi }{6}) + \binom{9}{7}(e^\frac{35\pi}{6}) (e^\frac{16\pi }{6}) +9 (e^\frac{40\pi}{6}) (e^\frac{8\pi }{6}) + (e^\frac{45\pi}{6}) \\ = (e^\frac{72\pi }{6}) + 9 (e^\frac{69\pi }{6}) + \binom{9}{2}(e^\frac{66\pi}{6}) + \binom{9}{3}(e^\frac{63\pi}{6}) + \binom{9}{4}(e^\frac{60\pi}{6}) + \binom{9}{5}(e^\frac{57\pi }{6}) + \binom{9}{6}(e^\frac{54\pi}{6}) + \binom{9}{7}(e^\frac{51\pi}{6}) +9 (e^\frac{48\pi}{6}) + (e^\frac{45\pi}{6}) \\ = (e^\frac{24\pi }{2}) + 9 (e^\frac{23\pi }{2}) + \binom{9}{2}(e^\frac{22\pi}{2}) + \binom{9}{3}(e^\frac{21\pi}{2}) + \binom{9}{4}(e^\frac{20\pi}{2}) + \binom{9}{5}(e^\frac{19\pi }{2}) + \binom{9}{6}(e^\frac{18\pi}{2}) + \binom{9}{7}(e^\frac{17\pi}{2}) +9 (e^\frac{16\pi}{2}) + (e^\frac{15\pi}{2}) \\\)

LaTex

\displaystyle (e^\frac{5\pi}{6} + e^\frac{4\pi}{3})^9\\
=\displaystyle \sum_{n=0}^9    \binom{9}{n}(e^\frac{5\pi\; n}{6})  (e^\frac{8\pi (9-n)}{6})\quad \\
= \binom{9}{0}(e^\frac{5\pi\; 0}{6})  (e^\frac{8\pi (9)}{6})
+ \binom{9}{1}(e^\frac{5\pi\; 1}{6})  (e^\frac{8\pi (8)}{6})
+ \binom{9}{2}(e^\frac{5\pi\; 2}{6})  (e^\frac{8\pi (7)}{6})
+ \binom{9}{3}(e^\frac{5\pi\; 3}{6})  (e^\frac{8\pi (6)}{6})
+ \binom{9}{4}(e^\frac{5\pi\; 4}{6})  (e^\frac{8\pi (5)}{6})
+ \binom{9}{5}(e^\frac{5\pi\; 5}{6})  (e^\frac{8\pi (4)}{6})
+ \binom{9}{6}(e^\frac{5\pi\; 6}{6})  (e^\frac{8\pi (3)}{6})
+ \binom{9}{7}(e^\frac{5\pi\; 7}{6})  (e^\frac{8\pi (2)}{6})
+ \binom{9}{8}(e^\frac{5\pi\; 8}{6})  (e^\frac{8\pi (1)}{6})
+ \binom{9}{9}(e^\frac{5\pi\; 9}{6})  (e^\frac{8\pi (0)}{6})\\

= (e^\frac{72\pi }{6})
+ 9(e^\frac{5\pi}{6})  (e^\frac{64\pi }{6})
+ \binom{9}{2}(e^\frac{10\pi}{6})  (e^\frac{56\pi }{6})
+ \binom{9}{3}(e^\frac{15\pi}{6})  (e^\frac{48\pi }{6})
+ \binom{9}{4}(e^\frac{20\pi}{6})  (e^\frac{40\pi }{6})
+ \binom{9}{5}(e^\frac{25\pi }{6})  (e^\frac{32\pi }{6})
+ \binom{9}{6}(e^\frac{30\pi}{6})  (e^\frac{24\pi }{6})
+ \binom{9}{7}(e^\frac{35\pi}{6})  (e^\frac{16\pi }{6})
+9 (e^\frac{40\pi}{6})  (e^\frac{8\pi }{6})
+ (e^\frac{45\pi}{6}) \\

= (e^\frac{72\pi }{6})
+ 9 (e^\frac{69\pi }{6})
+ \binom{9}{2}(e^\frac{66\pi}{6})  
+ \binom{9}{3}(e^\frac{63\pi}{6})  
+ \binom{9}{4}(e^\frac{60\pi}{6})  
+ \binom{9}{5}(e^\frac{57\pi }{6})  
+ \binom{9}{6}(e^\frac{54\pi}{6})  
+ \binom{9}{7}(e^\frac{51\pi}{6})  
+9 (e^\frac{48\pi}{6}) 
+ (e^\frac{45\pi}{6}) \\