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first die:  any number
second die:  prob 5/6 that it is different from first
third die:      prob  4/6 that it is different from first two

So, the probability they are all different=6/6 x 5/6 x 4/6 =120/216

So, the probability that at least 2 of them show the same face is:

1 - 120/216 = 96 / 216 =44.44%

Let the number of  red balls  = x     

So we have that

[x / 10] * [ (x - 1) / 9 ]  = 1/15     simplify

x(x - 1) / 90   = 1/15       multiply both sides by  90

x ( x - 1)   = 6   

x^2 - x -  6  = 0     factor

(x -3) ( x + 2)  = 0    set each facotr to 0 and solve for x and we get that

x = 3    or  x  = -2       reject the second answer 

So...there are 3  red balls

Proof :  First Draw Probability of  Red  * Second  Draw Probability of  Red  =

                   3 / 10   *  2 / 9   =  6 / 90    = 1 / 15

As follows:

.

yes, this is exactly what I'm looking for, tysm

Hope this helps:

We can really construct this scenario precisely: the first toss can be anything, then the second toss can be all but what the first toss was, the third toss can be all but what the second toss was, etc., up through the ninth toss. The tenth toss, though, must be exactly what the ninth toss was. So, the probability is the product of the probabilities that the second to ninth tosses are all different than the previous toss and the tenth is the same of the ninth: 1 * 5/6 * 5/6 ... * 5/6 *1/6 = 5^8 / 6^9 = 0.039.

I think this is what you are  looking for, but I'm not sure... I'll give it my  best shot  using something known as  the "sum of differences "

4  16    42    88     160

  12   26   46   72

      14   20   26

           6     6

              0

Using the polynomial  form  p(x)  = ax^3 + bx^2 + cx + d , we can  solve this system  of equations  for a,b, c and  d

a  +  b +  c +  d  =  4         [ for x  = 1, the first term]

8a + 4b + 2c + d  = 16        [for x  = 2, the second term]

27a + 9b + 3c + d  = 42      [for  x  = 3, the third term] 

64a + 16b + 4c + d  = 88     [for x  = 4, the fourth term ]

 The solving process is a little lengthy [ but not difficult ] ...I used WolframAlpha to find the solutions

a = 1, b  = 1,  c  = 2  and  d  = 0

So....the generating polynomial p(x) =   [ a_n ]  =    x^3 +  x^2 + 2x

Note that the first  five  terms are as above .....https://www.wolframalpha.com/input/?i=x%5E3+%2B+x%5E2+%2B+2x,++for++x++%3D+1,2,+3,+4,5

I hope that helped....if this is not what you want.....check back.....our member "heureka"  is  pretty knowledgeable about this sort of thing

OK, Coldplay....!!!

This triangle is isoceles.....the shortest angle bisector  will  be the one that bisescts angle "A"....it will be the altitude of the triangle

We can find the length of this bisector using the Pythagorean Theorem

Length  =   √ [ 14^2 -  13^2 ]   =  √ [196 - 169 ] =  √27  =  3√3 units

Here's a pic with AD, BE  and CF  being the bisectors.....as you can see,  AD  is the shortest bisector

This is from book VII of Euclid’s Elements.

CPhill, I thought you had your own personalized copy signed by Euclid himself.

Thanks so much Chris!

Actually....this could be done more simply

Draw  AD....now....we have ΔACD  with  angle ACD  = 150°

And since CD  = CA, then angle CAD  and And ADC  are equal

So angle CAD = [180 -150 ] / 2  = 30/2  =  15°

And if we draw, AE, by the same process,  in Δ ABE, angle EAB = 15°

So CAB  = 60°  =  angle CAD + angle DAE + angle EAB

 60   =  15  + angle DAE + 15

60  = 30 + angle DAE

30° =  angle DAE

That's pretty nifty, Melody... I don't think I am  familiar with this  !!!

x  = k / y   .....hold  k  constant

If y  is increased  by 25%   we have that

x  = k /  [ (5/4)y  ]     simplify

x = (4/5)k / y

Then  x  is 4/5 ths  =  80%  of its orginal value  ( it  decreases  by 20% ) 

If the product of all 6 rolls is, odd...EVERY roll must be odd, since any even roll - or multiple even rolls -  in the 6  will  result in an even product

We have a (1/2)  probability of an odd roll on each roll....so....the probability of  an odd product on  6  rolls  is :

(1/2)^6   =  1 / 64

ΔABE  ~ Δ BEC   which implies that

AB/ BE = BE / EC

And  ΔABE   is a 30 - 60 - 90 right triangle....AE  = 24  and BE  = 1/2 of this  = 12

So

24 / 12  = 12 / EC

12 /6  = 12 / EC

EC   =  CE  = 6

The answer is 12/25, because he can either be walked on the first appearence or the second. In both of those cases, there is a 2/5 probability that he is walked, and then a 3/5 probability that he is not walked. So, the final answer is (3/5*2/5)+(3/5*2/5) = 12/25.

Thanks, azsun...here's one more way.....

We have that   6 + b + c  = 18   ⇒   c = [18 - 6  - b]  =  [12  - b]

So...using the  Pythagorean Theorem

6^2  + b^2  = (12 - b)^2

36 + b^2  = 144 - 24b + b^2

24b - 108  = 0

24b  = 108

2b  = 9

b  = 9/2  = 4.5

So...c  =  12  - 4.5  =  7.5

Mmmm...I assumed the second way....but I see your point, Guest  !!!

Let's assume your first way is the  correct one...so we have

1 2 3 4 5 6 7 8 9 10 11 12 13

12 11 10  9  8  7  6  5 4 3 2 1

2 3 4 5 6 7 8 9 10 11 12 13

12 11 10  9  8  7  6  5  4 3  2  1

So...the first  row contains 13 total digits...and every othe row will contain 12 digits

Subtracting 13 from 5000 we get 4987  digits remaiining

So   4987 / 12 = 415 * 12 + 7

So....this will be the 7 digit on the 416th row after the first one

Which will be the row  2 3 4 5 6 7 8 9 10 11 12 13

So...the 5000th digit will still   = 8  

Just wan tto try this one myself....!!!

The triangle will be a45-45-90 right triangle

Call the length of its legs, x

So...the area of the black triangle is  (1/2) (product of the leg lengths)  = (1/2)x^2

We also have two  "white" rectangles  each with an area  of  x ( √6 - x  )  = 2x (√6 - x)

And we have a "white" square  with an area  of (√6 - x)^2

Since the areas are equal...we have that

(1/2)x^2  = 2x(√6 - x)  + (√6 - x)^2

(1/2)x^2  = 2√6x - 2x^2  + 6  - 2√6x + x^2

(1/2)x^2  = -x^2 + 6

(3/.2)x^2  - 6   = 0

3x^2 - 12  = 0

3x^2  = 12

x^2  = 4

x =2

So...when  A  is folded over.....it will  be √ [ 2^2 + 2^2 ] = √8  =  2√2 cm  from it's original position

Does he count like THIS?:

1  2  3  4  5  6  7  8  9  10  11 12  13  12  11 10  9  8  7  6  5  4  3  2  1  2  3  4  5  ......

or like THIS: ?

1 2  3  4  5  6  7  8  9 10 11 12  13   13  12  11 10  9  8  7  6  5  4  3  2  1  1  2  3  4  5  ....

It SOUNDS like the former.....

So you would START with 13 up the 12 DOWN, then 12 UP then 12 DOWN

Note that we  have a  cycle of  26 digits - 13 up and 13  down

So

5000 mod 26  =  8

So 8 will be the 5000th digit

Proof

192  * 26  +  8    =  5000

Find The Equation Of The Line That Passes Through The Points (-a,a) And (-3a,3a).

x = 1 and y = 1

y = k / x, where k=constant

k = xy = 1

1.25 = k / x

1.25 = 1 / x

1.25x = 1

x = 1 / 1.25

x = 0.8 - so, if y increases by 25% x decreases by 20%.

lim_(x->∞) (sqrt(4 x^2 + x) + (sqrt(9 x^2 - x) - 5 x)) = 1/12

\( \displaystyle\lim_{x\rightarrow \infty}\;\;[\sqrt{4 x^2 + x}+ \sqrt{9 x^2 - x} - 5 x]= \frac{1}{12}\)

There is nothing to solve.  This is either true or not true and it is not true

Have you written the question incorrectly?