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12-(8-2)=12-6=6

If there were 3 skiers on plane of 17 and 4 people on the plane died of a crash, what is the chance that all 3 skiers survive?

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This probability is trickier than it looks.

Four (4) will die and at most only three (3) can be skiers.

By deduction one (1) will die and not be a skier.

Now 16 remain, three (3) of whom are skiers.

From this calculate the probability of selecting (Z) correct out of (R) draws from (N) numbers. (Z) (in this case) defines the probability of a skier dying.

Probability= (R!/(Z!*(R-Z)!) * (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)/(N!/((R!)*((N-R)!)))

 $${Probability =}\frac{\frac{R!}{Z!*(R-Z)!} * \frac{(N-R)!}{((N-R)-(R-Z))!*(R-Z)! }}{\frac{N!}{R!*(N-R)!}}$$

 Column ID’s: A= (R!)/(Z!(R-Z)!)

                    B= (N-R)!/(((N-R)-(R-Z))!*(R-Z)!)

                    C= N!/((R!)*((N-R)!))

                    D= Probability of (Z) skiers dying.

                    E= 1/Probability

N     R     Z     A     B      C                   D                            E

16   3      3     1     1      560     0.001785714285714     560.0000000000

16   3      2     3    13     560     0.069642857142857       14.3589743590

16   3      1     3    78     560     0.417857142857143        2.3931623932

16   3      0     1  286     560     0.510714285714286        1.9580419580

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Probability of zero skiers dying ~ 0.511 (51.1%)

(Professional help provided by Francis and Frances)

by: Someone Who Knows Everything

Okay, thanks Heureka,

I think that I have got it.  

hi Melody,

{rll...} is the alignment.

1. Field r (right-alignment) bevor first "&"

2. Field l (left-alignment) after first "&"

3. Field l (left-alignment) after second "&"

and so on...

r = right alignment

c = center alignment

l = left alignment

I hope, I could help you

heureka

\begin{array}{rll} 2x-8&=&6-5x\\ 7x&=&14\\ x&=&2 \end{array}

$$\begin{array}{rll}

2x-8&=&6-5x\\
7x&=&14\\
x&=&2
\end{array}$$

From: Someone Who Knows Everything ... … …

I am almost humbled to be in this land of Camel Lot… Never the less, to the questions and proofs at hand or foot …

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… assuming that you do indeed know everything - you may still tell lies - and if you tell lies then maybe you really do not know everything - It does sound like a circular arguement - We need Sir Cumference here to arbitrate! :by Melody

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Sir Cumference, (assuming he still has his head) would argue that I know “everything” about lies.

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 ...I'm busy at the moment trying to extricate a Roman zero from a boulder. It's a hyperbolic problem...........and that' s no hyperbole.. :by CPhill

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 Yes, it is a hyperbolic problem. We need a set of hyperbolic equations and a set of hyperbolic relations such that they converge at 42. The hyperbolic equations should contain most of the truth and the hyperbolic relations should contain most of the lies (and maybe contain the “Roman Zero”).

Now for the circular argument:

If our collective knowledge is the diameter of a circle, then is the circumference of the circle a measure of our ignorance?

Basic descriptive principle: as the diameter of knowledge increases, the circumference of ignorance increases by a factor of Pi.

If this is true, then the knowledge we gain increases our ignorance by a factor of Pi. This would mean I am the most ignorant person in the world.¹ By the same measure, someone who knows nothing is the least ignorant. ²

Conceivably the circumference does not measure ignorance. It probably measures a border of ignorance – a border of what we do not know. The tangents of the circle are where we are aware of what we do not know. Farther out is where we are unaware of what we do not know.

A circle has an area. What does this define? Wisdom? If so, then wisdom increases as ½ the diameter squared times Pi. This seems too fast a rate.

This resonates like a paradox worthy of Sir Cumference.

Increase its complexity by defining this in three dimensions: If our collective knowledge is the diameter of a sphere, then … … …

This now resonates like a paradox worthy of Sir Phere, the progeny of Sir Cumference. However, that is another ball of wax. Anyone want to play ball?

If so, you will have to wait. I was doing this while skiing and I just received a call to investigate a plane crash.

Strange fixation: Someone keeps rolling a big bolder up the ski slope, while another keeps looking for something (or nothing) inside. He has three sets of sunglasses. Strange world we live in!

In the mean time, think outside the circle or sphere: question the basics. You may look like a fool, but, if you do not, you will be … for a lifetime. Believe me, I know, because I know everything. (There were a couple of people here, who did that once, …which is part of the reason I know everything).

Some of you now probably “know” something you did not know before. Those that do know are intelligent enough to know that some information is more valuable when it is esoteric.

By: Someone Who Knows Everything

Notes:

1) (And among the most arrogant).

2) (Not necessarily the least arrogant).

hi Heureka,

I have only done this type of setting out with equarray.

but i haven't been able to get equarray to work on this forum.

Thankyou for showing me how I can do it with array.

I don't quite understand what {rll}stands for.

Can you explain please?

I think I will try equarray again and see if I can get it to work. I tried again - it still doesn't work.

Hi Anonymous,

If you are a conscientious objector we can respect that.

If you post a lot why don't you give yourself a unofficial username - you can put it at the end of all your posts and we will slowly learn who you are.

We would really like you to be a proper 'member' of the forum even if it is unofficial.   

No it is not

5=11+2(x-1)    x?

$$\begin{array}{rll}
5&=&11+2(x-1)\quad | \quad -11\\
5-11&=& 2(x-1)\\
-6&=& 2(x-1)\quad | \quad :2\\
-\frac{6}{2}&=& (x-1)\\
-3&=& x-1\quad | \quad +1\\
-3+1&=& x\\
-2&=& x\\
\end{array} \\
\boxed{x=-2}\\$$

Latex Code:

\begin{array}{rll}
5&=&11+2(x-1)\quad | \quad -11\\
5-11&=& 2(x-1)\\
-6&=& 2(x-1)\quad | \quad :2\\
-\frac{6}{2}&=& (x-1)\\
-3&=& x-1\quad | \quad +1\\
-3+1&=& x\\
-2&=& x\\
\end{array} \\
\boxed{x=-2}\\

Proof:

$$\begin{array}{rll}
5&=&11+2*(\textcolor[rgb]{1,0,0}{-2}-1)\\
5&=&11+2*(-3)\\
5&=&11-2*3\\
5&=&11-6\\
5&=&5\\
\end{array} \\$$

Latex code:

\begin{array}{rll}
5&=&11+2*(\textcolor[rgb]{1,0,0}{-2}-1)\\
5&=&11+2*(-3)\\
5&=&11-2*3\\
5&=&11-6\\
5&=&5\\
\end{array} \\

$$5=11+2(x-1)\\
5-11=11+2(x-1)-11\\
-6=2x-2\\
-6+2=2x-2+2\\
-4=2x\\
-4/2=x\\
x=-2$$

COOL!!! 

That's brilliant!  Thanks Bertie.  I'll give you another 3 points.  For sure!!!

AAAAAAAAAAAAAAARRRRRGGGHHH!

BY THE BEARD OF ZEUS!

THOU SHALT NOT TORMENT THY HOLY LAWS!

THOU SHALL FEAR THE WRATH OF MATH!

I CURSE THEE WITH UNBREAKABLE BONDS!

$$P=NP / P \neq NP$$

LOL!!!

I'd give you 6 points for that one......if I could!!!!

But what if the plane was carrying 4 people that died in a crash in coffins in the cargo room. Then the plane might not have crashed at all and the skiers survive 

Unless the skiers died in a skiing accident off course 

@@ End of Day Wrap - Wednesday 21/5/14   Sydney, Australia   Time 19:35

Hi Everybody,

Great answers were posted today by CPhill, problem, admin, zegroes, Bertie, Reinout-g, GoldenLeaf, Heureka and Kitty

Zegroes also reported a problem with the message centre by posting it on the forum.  This was a great idea.  Admin saw the post and responded quite quickly.  Thanks to both of you.

There were plenty of questions but I don't think any really stood our for their mathematical content.

These ones may provide some entertainment.

The fun continued with this first one:

Have you heard of the story of the red indian chief who had three squaws (wives) ?

They were allowed to choose the bedding on which they slept, each according to her own preference. The first chose to sleep on a hippopotamus skin, the second on a lion skin and the third on a tiger skin. (And yes I know what you are going to say, but this was long ago when there were hippopotami and tigers in North America.)

In the fullness of time, the first wife bore triplets, the second wife had twins, but the third could only manage a single baby.

The moral of the story is that "The squaw on the hippopotamus is equal to the sum of the squaws on the other two hides".

That's scarecrows for you!  They just cannot be trusted 

CD

And I'm still looking for the Roman Numeral "0".......eh, never mind, it's an "inside" joke here on the forum.....

Here is an update on the probability puzzles!
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Easier one (hasn't been answered yet): 

The fabulous game host Steve offer you a bag with 5001 pearls of which 2501 are white and 2500 are black. He asks you to draw as many pearls from the bag as you desire without looking. Steve then tells you how many black and white pearls you drew. If there are equally as much white as black pearls in the draw you gain €10 for each pearl you drew. How many pearls should you draw?

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Hard one: 

All right, so here's a classic one.

Jason and Tyrell are in a classroom of 23 students.

At some point Tyrell finds out that some of his classmates have the same birthday.

In surprise he exclaims; 'what a surprise!'.

Jason however, says it is more likely for two or more people to have the same birthday in their class than it is for noone to have the same birthday.

Assume any birthday is equally likely to another and pretend leap years don't exist. (it makes it a lot easier)

What are the odds of two or more people sharing their birthday?

There are just about always two methods for dealing with problems like this, a counting method and a probability method, (and often one is more convenient than the other.)

The counting method has been used for both of the correct answers so far,  so here's the probability method.

Suppose that all 17 were injured, some of them critically, and that 4 of them subsequently die, (at intervals of several minutes say). The probability that the first to die is a non-skier is 14/17, that the second third and fourth to die are also non-skiers 13/16, 12/15 and 11/14 respectively. The required probability will be the product of these, 24024/57120 = 143/340.

Okay, so I think Creepercosmo was so close to the answer that I'm gonna give him 5 points...

Here's how it can be done.

Let's say there are indeed 4 options.

1. I confess, my friend does too and we both spend 10 years in jail

2. I confess, my friend does not and he spends 20 years on his own

3. I do not confess, my friend does and I spend 20 years on my own

4. I do not confess, my friend doesn't either and we both spend 3 years

Let's see how many suffering points each situation gives me;

1. s = 2*10 + 10 = 30

2. s = 2*0 + 20 = 20

3. s = 2*20 + 0 = 40

4. s = 2*3 + 3 = 9

Let p be the estimated probability that your partner confesses.

Then the expected amount of suffering is

If I confess;

E(S|C) = 30p+20(1-p) = 10p+20

If I do not confess;

E(S|NC) = 40p+9(1-p) = 31p+9

We want to find a value for p where you're indifferent between confessing or not confessing.

Therefore

$$10p+20 = 31p+9 \Rightarrow 21p = 11 => p = \frac{11}{21} = 0.5238$$

Now the slope of 31p+9 is obviously higher than that of 10p+20 (and we want to minimize the amount of suffering)

Therefore if $$p>0.5238$$ You'd choose to confess and

if $$p<0.5238$$ You'd choose not to confess.

Now obviously, I highly doubt someone would say his friend has a 52.39% chance of confessing, so well done Creepercosmo! 

I'm going to think of a new one 

And for those who paid close attention........

In "The Wizard of Oz,"   the scarecrow - after getting a "brain" from the Wizard - states:

"The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side."

It sounds convincing when he says it......but it's wrong.......