Sociopath Community

Golden_Eagle said: 

Will we get f/n=60/400=0.15 again  ? 

I'm guessing it would have to be done before we can write something like f/n=45/400, whatever the result would be. If the experiment is identical then must have a stable frequency.

If i rolled two die again I probably would not get 60/400 and therefore the probability of obtaining a sum of 7 from a my dice roll probably is not 0.15.

This is true regardless of exact ideal conditions during a repeated experiment.

Think in terms of a coin toss. If I flipped a coin 1 time and it came up heads then f/n = 1/1 = 1 and 1 implies that the coin should come up heads every single time but we know that is silly. If the coin is fair then we know the probability of heads is 0.5. The greater the number N the closer the actual probability of heads will be 0.5.

Is that with identical conditions as well ?

It depends on what is meant by 'identical'.

We could create an idealized version of the coin flip in which all the physics are the exact same. Consider flipping a coin in a vacuum, there is no air or any other particles to interact with the coin, you flip it with the same amount of force and angle, etc. Under said conditions you would be able to predict exact results, it would no longer be a question of statistics and probability but instead classical mechanics.

In statistics and probability of games we assume identical conditions to mean the same level of uncertainty is present in the experiment. Below I simulate coin flips using a computer, in it I assume the same level of uncertainty for each coin flip by using a random number generator that assumes the coin flips are normally distributed. 

Look how the relative frequency (probability) of heads changes as we increase the number N.

N = 10

N = 50

N= 10000

I wouldn't go with a pair of seven cause I'm not sure how that works with 6 sided dice. Maybe with 4 dice, but that would at least decrease the odds when rolling a pair of 7 in a single roll, but that shouldn't matter, it's the result that determins the relative frequency.

Maybe I don't understand that part, so I'll settle for a pair of 6. Probably scaled down from the OP model. I should probably use 1 dice for starters.

This is good for short term memory, but that's the best I can do for now. 

P₆ = {(5,1),(4,2),(3,3),(1,5),(2,4)}, so there are actually only 5 ways to get a sum of 6 while there is 6 ways to get a sum of 7, hence betting on seven is technically better.

Alice I don't get it. ( Those graphs make life easier ) But the part I don't get...

I am not Alice.

.

Wait I see, had to count them.  Probability 6 P₆={(5,1),(4,2),(3,3),(1,5),(2,4)} each adds up to 6. And each are the possible outcomes. I was thinking 6 sided dice, but this isn't the same.

Oh wait, the B was assigned, but that isn't the case this time. ( I'm writing my thoughts to show you how amature I am are getting this. I'm not quick at it. )

Probability is not easy by any means, it is, however, incredibly rewarding.

There is a trade off between adding more dice because despite there now being a greater number of ways to obtain any given sum there is also a much larger set of possible outcomes as a whole. I won;t do the calculations now but I wonder how probabilities of different sums of pairs scales with increasing dice.

Got it. 

In the previous result of f/n=60/400=0.15 ( let's say stable ) would we be able to simply use addition and make it 

f/n=240/1600=0.60 ? Or is that cheating ? ( I have my doubt about this, maybe there's something I don't see )

On its own that is cheating and cheating in probability will leave you ignorant given the outcome is making incorrect inferences.

You can do a similar process however via computational simulation, much like the one I created above. A simulation is cheating in the sense that it is just made up and built completely on our own assumptions about how we believe something should behave given uncertainty.

For instance, in the simulation of coin flips I write in the assumption that coin flips are normally distributed, that is they follow the bell curve. How valid is that assumption, it depends on my goal. 

Typically the goal is to fit a simulation to a real world systems behavior. We observe something uncertain in real life and then create a computer simulation of that things behavior up to a point so that we can understand its uncertainty to a great enough fidelity that we can predict future outcomes or other properties of that things behavior.

This is limited by how complex the thing we are simulating is and a long the way we must always consider how much trust we can appropriately allocate to our simulation.

it's the result that determins the relative frequency.

if we consider f/N where it is the number of occurrences of an even and the number of samples that determines the relative frequency.

Infinite samples (N -> ∞) is what gives us our absolute probability of an event, that is a coin flip is 0.5 heads and 0.5 tales.

 Is there a machine that can say, flip a coin with the exact identical conditions ( down to the coin's exact starting point) and achieve the same result at various frequencies ? ?

This took long to write, so much thinking. It's easier to get it done faster eh ?

 I believe for coin flips that is very possible and I know that there are magicians who are capable of flipping heads almost every time. Coin flipping is simple and as such to manipulate it is simple. Black Jack is a great deal more complicated but still relatively simple, and though you cannot predict every hand with absolute certainty you can predict a set of hands x% of the time with absolute certainty which is why counting cards is actually profitable. A casino can increase the complexity of a black jack game by adding additional card decks and rules for betting. With enough complexity counting cards becomes impossible and the odds are back in the houses favor.

Adelaide said: 

Golden_Eagle said: 

Will we get f/n=60/400=0.15 again  ? 

I'm guessing it would have to be done before we can write something like f/n=45/400, whatever the result would be. If the experiment is identical then must have a stable frequency.

If i rolled two die again I probably would not get 60/400 and therefore the probability of obtaining a sum of 7 from a my dice roll probably is not 0.15.

This is true regardless of exact ideal conditions during a repeated experiment.

Think in terms of a coin toss. If I flipped a coin 1 time and it came up heads then f/n = 1/1 = 1 and 1 implies that the coin should come up heads every single time but we know that is silly. If the coin is fair then we know the probability of heads is 0.5. The greater the number N the closer the actual probability of heads will be 0.5.

Is that with identical conditions as well ?

It depends on what is meant by 'identical'.

We could create an idealized version of the coin flip in which all the physics are the exact same. Consider flipping a coin in a vacuum, there is no air or any other particles to interact with the coin, you flip it with the same amount of force and angle, etc. Under said conditions you would be able to predict exact results, it would no longer be a question of statistics and probability but instead classical mechanics.

In statistics and probability of games we assume identical conditions to mean the same level of uncertainty is present in the experiment. Below I simulate coin flips using a computer, in it I assume the same level of uncertainty for each coin flip by using a random number generator that assumes the coin flips are normally distributed. 

Vacuum then with no drag or disturbances. You've answered my question.

Look how the relative frequency (probability) of heads changes as we increase the number N.

N = 10

N = 50

 

I wouldn't go with a pair of seven cause I'm not sure how that works with 6 sided dice. Maybe with 4 dice, but that would at least decrease the odds when rolling a pair of 7 in a single roll, but that shouldn't matter, it's the result that determins the relative frequency.

Maybe I don't understand that part, so I'll settle for a pair of 6. Probably scaled down from the OP model. I should probably use 1 dice for starters.

This is good for short term memory, but that's the best I can do for now. 

P₆ = {(5,1),(4,2),(3,3),(1,5),(2,4)}, so there are actually only 5 ways to get a sum of 6 while there is 6 ways to get a sum of 7, hence betting on seven is technically better.

Alice I don't get it. ( Those graphs make life easier ) But the part I don't get...

I am not Alice.

Koloss.

I should probably bring her here, she's more algebra but can probably relate.

.

Wait I see, had to count them.  Probability 6 P₆={(5,1),(4,2),(3,3),(1,5),(2,4)} each adds up to 6. And each are the possible outcomes. I was thinking 6 sided dice, but this isn't the same.

Oh wait, the B was assigned, but that isn't the case this time. ( I'm writing my thoughts to show you how amature I am are getting this. I'm not quick at it. )

Probability is not easy by any means, it is, however, incredibly rewarding.

I'm quite enjoying this actually. It would be funny implement it and somehow it works accordingly, even if it's close.

There is a trade off between adding more dice because despite there now being a greater number of ways to obtain any given sum there is also a much larger set of possible outcomes as a whole. I won;t do the calculations now but I wonder how probabilities of different sums of pairs scales with increasing dice.

Got it. 

In the previous result of f/n=60/400=0.15 ( let's say stable ) would we be able to simply use addition and make it 

f/n=240/1600=0.60 ? Or is that cheating ? ( I have my doubt about this, maybe there's something I don't see )

On its own that is cheating and cheating in probability will leave you ignorant given the outcome is making incorrect inferences.

You can do a similar process however via computational simulation, much like the one I created above. A simulation is cheating in the sense that it is just made up and built completely on our own assumptions about how we believe something should behave given uncertainty.

For instance, in the simulation of coin flips I write in the assumption that coin flips are normally distributed, that is they follow the bell curve. How valid is that assumption, it depends on my goal. 

Typically the goal is to fit a simulation to a real world systems behavior. We observe something uncertain in real life and then create a computer simulation of that things behavior up to a point so that we can understand its uncertainty to a great enough fidelity that we can predict future outcomes or other properties of that things behavior.

This is limited by how complex the thing we are simulating is and a long the way we must always consider how much trust we can appropriately allocate to our simulation.

This is the kind of stuff they can use to merge robotics with the physical world. Though I have no idea how fast or how long it would take for, say, machine learning to be able to operate with it in real time. I'm familiar with 3D simulation, but I'm more of a software user than a coder. I do have 27 RX 580 8GB AMD cards, but I have no idea how to do any of that Ai stuff, other than donate hash power, and of course mine. I recall accuracy takes more time to render out.

it's the result that determins the relative frequency.

if we consider f/N where it is the number of occurrences of an even and the number of samples that determines the relative frequency.

Infinite samples (N -> ∞) is what gives us our absolute probability of an event, that is a coin flip is 0.5 heads and 0.5 tales.

 Is there a machine that can say, flip a coin with the exact identical conditions ( down to the coin's exact starting point) and achieve the same result at various frequencies ? ?

This took long to write, so much thinking. It's easier to get it done faster eh ?

 I believe for coin flips that is very possible and I know that there are magicians who are capable of flipping heads almost every time. Coin flipping is simple and as such to manipulate it is simple. Black Jack is a great deal more complicated but still relatively simple, and though you cannot predict every hand with absolute certainty you can predict a set of hands x% of the time with absolute certainty which is why counting cards is actually profitable. A casino can increase the complexity of a black jack game by adding additional card decks and rules for betting. With enough complexity counting cards becomes impossible and the odds are back in the houses favor.

 No idea how that can be done. I've seen footage of people getting kicked out for counting cards though. Counting after a shuffle sounds pretty insane.

Golden_Eagle said: 

Koloss.

I should probably bring her here, she's more algebra but can probably relate.

I am not Koloss.

I am down with anyone who appreciates numbers and computers.

There is a trade off between adding more dice because despite there now being a greater number of ways to obtain any given sum there is also a much larger set of possible outcomes as a whole. I won;t do the calculations now but I wonder how probabilities of different sums of pairs scales with increasing dice.

Got it. 

In the previous result of f/n=60/400=0.15 ( let's say stable ) would we be able to simply use addition and make it 

f/n=240/1600=0.60 ? Or is that cheating ? ( I have my doubt about this, maybe there's something I don't see )

On its own that is cheating and cheating in probability will leave you ignorant given the outcome is making incorrect inferences.

You can do a similar process however via computational simulation, much like the one I created above. A simulation is cheating in the sense that it is just made up and built completely on our own assumptions about how we believe something should behave given uncertainty.

For instance, in the simulation of coin flips I write in the assumption that coin flips are normally distributed, that is they follow the bell curve. How valid is that assumption, it depends on my goal. 

Typically the goal is to fit a simulation to a real world systems behavior. We observe something uncertain in real life and then create a computer simulation of that things behavior up to a point so that we can understand its uncertainty to a great enough fidelity that we can predict future outcomes or other properties of that things behavior.

This is limited by how complex the thing we are simulating is and a long the way we must always consider how much trust we can appropriately allocate to our simulation.

This is the kind of stuff they can use to merge robotics with the physical world. Though I have no idea how fast or how long it would take for, say, machine learning to be able to operate with it in real time. I'm familiar with 3D simulation, but I'm more of a software user than a coder. I do have 27 RX 580 8GB AMD cards, but I have no idea how to do any of that Ai stuff, other than donate hash power, and of course mine. I recall accuracy takes more time to render out.

https://en.wikipedia.org/wiki/AIXI

This is considered the most optimal and universal solution to artificial general intelligence. It builds probabilistic models of its environment and then rates those models on their accuracy and simplicity. The main idea is that the program that is the shortest but can make the most accurate inferences is the best.

The whole paradigm is known as Compression = Intelligence.

It was proven incomputable by its own designer, so the fruitful path is a matter of approximation. 

Something similar is Godel Machine, https://en.wikipedia.org/wiki/G%C3%B6del_machine

it's the result that determins the relative frequency.

if we consider f/N where it is the number of occurrences of an even and the number of samples that determines the relative frequency.

Infinite samples (N -> ∞) is what gives us our absolute probability of an event, that is a coin flip is 0.5 heads and 0.5 tales.

 Is there a machine that can say, flip a coin with the exact identical conditions ( down to the coin's exact starting point) and achieve the same result at various frequencies ? ?

This took long to write, so much thinking. It's easier to get it done faster eh ?

 I believe for coin flips that is very possible and I know that there are magicians who are capable of flipping heads almost every time. Coin flipping is simple and as such to manipulate it is simple. Black Jack is a great deal more complicated but still relatively simple, and though you cannot predict every hand with absolute certainty you can predict a set of hands x% of the time with absolute certainty which is why counting cards is actually profitable. A casino can increase the complexity of a black jack game by adding additional card decks and rules for betting. With enough complexity counting cards becomes impossible and the odds are back in the houses favor.

 No idea how that can be done. I've seen footage of people getting kicked out for counting cards though. Counting after a shuffle sounds pretty insane.

I do not know the details of card counting but I do know that some time ago an AI figured out how to play black jack most optimally. Most optimally is defined as playing the perfect move for each given hand between itself and the dealer. The AI showed that if you play perfectly you will only win 49% of the time, so that is proof that black jack is a losing game and that 51% win rate for the casino is how they win all there money. 1% is all they need to run a profitable business.

Card counting gives you something like 53% win rate which turns the game in your favor. Mathematically it is method to gauge how random the deck is at any given moment, when it is less random you are willing to up your bets and and when it is more random you only bet the table minimum or don't play at all.

All of this was discovered by a guy named Edward Thorp.

https://en.wikipedia.org/wiki/Edward_O._Thorp#Computer-aided_research_in_blackjack

Solutions to section 1.2

1.2.1 Find the union C₁∪C₂ and the intersection C₁∩C₂ of the two sets C₁ and C₂ where

(a) C₁ = {0,1,2}, C₂={2,3,4}
C₁ ∪ C₂ = {0,1,2,3,4}
C₁ ∩ C₂ ={2}

(b) C₁ = {x: 0 < x < 2}, C₂ = {x: 1 <= x < 3}
C₁ ∪ C₂ = {0 < x < 3}
C₁ ∩ C₂ = {1 <= x < 2}

(c) C₁ = {(x,y): 0 < x < 2, 1 < y < 2}, C₂ = {(x,y): 1 < x < 3, 1 < y < 3}
C₁ ∪ C₂ = {(x,y): 0 < x < 3, 1 < y < 3}
C₁ ∩ C₂ = {(x,y): 1 < x < 2, 1 < y < 2}

1.2.2 Find the complement Cᶜ of the set C with respect to the space Ξ.

(a) Ξ = {x: 0 < x < 1}, C = {x: 5/8 < x < 1}
Cᶜ = {x: 0 < x < 5/8}

(b) Ξ = {(x,y,z): x² + y² + z² <= 1}, C = {(x,y,z): x² + y² + z² = 1}
Cᶜ = {(x,y,z): x² + y² + z² < 1}

(c) Ξ = {(x,y): |x| + |y| <= 2}, C = {(x,y): x² + y² < 2}
Cᶜ = {(x,y): |x| + |y| <= 2 <= x² + y²}

1.2.3 List all possible arrangements of the four letters m,a,r,and y. Let C₁ be the collection of the arrangements in which y is in the last position. Let C₂ be the collection of arrangements in which m is in the first position. Find the union and the intersection of C₁ and C₂.

mary, mray, mrya, myra, myar, mayr
amry, army, arym, ayrm, aymr, amyr
rmay, ramy, raym, ryam, ryma, rmya
ymar, yamr, yarm, yram, yrma, ymra

C₁ = {mary, mray, amry, army, rmay, ramy, ymar, yamr}
C₂ = {mary, mray, mrya, mrya, myar, mayr}

C₁ ∪ C₂ = {mary, mray, mrya, myra, myar, mayr, amry, army, rmay, ramy, ymar, yamr}
C₁ ∩ c₂ = {mary, mray}

1.2.6 Show that the following sequences of sets {Cₖ} are increasing then find lim(k->∞){Cₖ}

(a) Cₖ = {x: 1/k <= x <= 3 - 1/k} k=1,2,3,...
Cₖ₊₁ = {x: 1/(k+1) <= x <= 3 - 1/(k+1)}
We know that 1/(k+1) < 1/k and 3 - 1/k < 3-1/(k+1)
thus, 1/(k+1) < 1/k < x < 3-1/k < 3 - 1/(k+1)
Given x ∈ Cₖ₊₁ and Cₖ ⊂ Cₖ₊₁
{Cₖ} is increasing
As k -> ∞, 1/k -> 0
As k -> ∞, 3 - 1/k -> 3
Hence, lim(k->∞){Cₖ} = {x: 0 < x < 3)

(b) Cₖ = {(x,y): 1/k <= x² + y² <= 4 - 1/k}, k = 1,2,3,...
Cₖ₊₁ = {(x,y)L 1/(k+1) <= x² + y² <= 4 - 1/(k+1)}, k= 1,2,3,...
We know 1/(k+1) < 1/k and 4 - 1/k < 4 - 1/(k+1)
thus, 1/(k+1) < 1/k <= x² + y² <= 4 - 1/(k) < 4 - 1/(k+1)
given x ∈ Cₖ₊₁ and Cₖ ⊂ Cₖ₊₁
{Cₖ} is increasing
As k -> ∞, 1/k -> 0
As k -> ∞, 4 - 1/k -> 4
Hence, lim(k->∞){Cₖ} = {(x,y): 0 < x² + y² < 4)

1.2.7 Show that the following sequences of sets {Cₖ} are decreasing then find lim(k->∞){Cₖ}

(a) Cₖ = {x: 2 - 1/k < x <= 2}, k=1,2,3,...
Cₖ₊₁ = {x: 2 - 1/(k+1) < x <=2}, k=1,2,3,...
We know that 2 - 1/k < 2 - 1/(k=1)
thus, 2-1/k < 2 - 1/(k+1) < x <=2
Given x ∈ Cₖ₊₁ and Cₖ₊₁ ⊂ Cₖ
{Cₖ} is decreasing
As k->∞, 2 - 1/k -> 2
so 2 < x <= 2
lim(k->∞){Cₖ} = {2}

(b) Cₖ = {x: 2 < x <= 2+1/k}, k=1,2,3,...
Cₖ₊₁ = {x: 2 < x <= 2+1/(k+1)}, k=1,2,3,...
We know that 2 + 1/k > 2 + 1/(k+1)
thus, 2 < x <= 2+1/(k+1) < 2 + 1/k
Given x ∈ Cₖ₊₁ and Cₖ₊₁ ⊂ Cₖ
{Cₖ} is decreasing
We know that x - 2 <= 1/k
However, there exists a k such that x - 2 > 1/k
Hence, there exists a k such that x > 2 + 1/k
This is a contradiction
Therefore, lim(k->∞){Cₖ} = ∅

(c) Cₖ = {(x,y): 0 <= x² + y² <= 1/k}, k=1,2,3,...
Cₖ₊₁ = {(x,y): 0 <= x² + y² <= 1/(k+1)}, k = 1,2,3,...
We know that 1/k > 1/(k+1)
thus, 0 <= x² + y² <= 1/(k+1) <= 1/k
Given x ∈ Cₖ₊₁ and Cₖ₊₁ ⊂ Cₖ
{Cₖ} is decreasing
As k->∞, 1/k -> 0
So, 0 <= x² + y² <= 0
lim(k->∞){Cₖ} = {(0,0)}

1.2.8 For every one-dimensional set C, define the function Q(C) = ∑_c f(x), where f(x) = (2/3)(1/3)ˣ, x=0,1,2,3.... If C₁ = {x: x = 0,1,2,3} and C₂ = {x: x=0,1,2,...} find Q(C₁) and Q(C₂).

For C₁ we have ∑_C₁ f(x)
∑_C₁ f(x) = (2/3)(1/3)⁰ + (2/3)(1/3)¹ + (2/3)(1/3)² + (2/3)/(1/3)³
              = 80/81
            ~= 0.99

For C₂ we have ∑_C₂ f(x)
∑_C₂ f(x) = (2/3)(1/3)⁰ + (2/3)(1/3)¹ + (2/3)(1/3)² + (2/3)/(1/3)³ + . . . + (2/3)(1/3)ⁿ
As n->∞, f(x) -> 1
∑_C₂ f(x) = 1

1.2.9 For every one-dimensional set C for which the integral exists, let Q(C) = ∫_c f(x)dx where f9x) = 6x(1 - x),
0 < x < 1, otherwise let Q(C) be undefined. If C₁ = {x: 1/4 < x < 3/4}, C₂ = {11/2}, and C₃ = {x: 0 < x < 10}.
Find Q(C₁), Q(C₂), and Q(C₃)

∫ f(x)dx = ∫ 6x(x-1)
            = ∫ 6x²-6x
            = -2x³ + 3x²

For C₁ we have ∫_c₁f(x)dx
∫_c₁f(x)dx = (-2(3/4)³ + 3(3/4)²) - (-2(1/4)³ + 3(1/4)²)
               = 11/16
             ~= 0.6875

Given C₂ we have ∫_c₂f(x)dx
∫_c₂f(x)dx = (-2(1/2)³ + 3(1/2)²) - (-2(1/2)³ + 3(1/2)²)
               = 0

Given C₃ we have ∫_c₃f(x)dx
∫_c f(x)dx is only defined on 0 < x < 1, hence we only compute ∫_c₃f(x)dx within those bounds
∫_c₃f(x)dx = (-2(1)³ + 3(1)²) - (-2(0)³ + 3(0)²)
               = 3 - 2
               = 1

1.2.10 For every two-dimensional set C contained in R² for which the integral exists, let Q(C) = ∫∫_c(x² + y²)dxdy. If C₁ {(x,y): -1 <= x <=1, -1 <= y <= 1}, C₂ = {(x,y): -1 <= x = y <= 1}, C₃ = {(x,y): x² + y² <= 1}. Find Q(C₁), Q(C₂), and Q(C₃).

For C₁ we have ∫∫_c₁(x² + y²)dxdy
∫∫_c₁(x² + y²)dxdy = ∫_c₁(1/3 + y² - (-1/3 -y²))dy
                           = ∫_c₁(2y² + 2/3)dy
                           = (2/3 + 2/3) - (-2/3 -2/3)
                           = 8/3
                         ~= 2.67

For C₂ we have ∫∫_c₂(x² + y²)dxdy
∫∫_c₂(x² + y²)dxdy = 0

For C₃ we have ∫∫_c₃(x² + y²)dxdy
We know that x² + y² = r² (definition of circle)
We convert to polar coordinates so that, x = rcosθ and y = rsinθ
∫∫_c₃(r²cosθ + r²sin²θ)drdθ = ∫∫_{(r,θ): 0<r<1, 0<θ<2π}(r³)drdθ
                                         = π/2

1.2.11 Let C denote the set of points that are interior to, or on the boundary of, a square with opposite vertices at the point (0,0) and (1,1). Let Q(C) = ∫∫_Cdydx.

(a) If c ⊂ C is the set {(x,y): 0 < x < y < 1}, compute Q(c).
∫∫_{{(x,y): 0 < x < y < 1}dydx = ∫₍₀₁₎ ∫₍ₓ₁₎dydx
                                            = ∫₍₀₁₎(1 - x)dx
                                            = 1 - 1/2
                                            = 1/2

(b) If c ⊂ C is the set {(x,y): 0 < x = y < 1}, compute Q(c).
∫∫_{{(x,y): 0 < x = y < 1}dydx = 0

(c) if c ⊂ C is the set {(x,y): 0 < x/2 <= y <= 3x/2 <1}, compute Q(c).
∫∫_{(x,y): 0 < x/2 <= y <= 3x/2 <1} = ∫_(0,2/3) ∫_(x/2,3x/2) dydx
                                                    = ∫_(0,2/3) (3x/2 - x/2)dx
                                                    = ∫_(0,2/3) x dx
                                                    = (2/3)²/2 - (0)²/2
                                                    = 2/9

1.2.12 Let C be the set of points interior to or on the boundary of a cube with edge length 1. Moreover, say that the cube is in the first octant with on vertex at the point (0,0,0) and an opposite vertex at the point (1,1,1). Let Q(c) = ∫∫∫_c dxdydz

(a) If c ⊂ C is the set {(x,y): 0 < x < y < z < 1} compute Q(c).
∫∫∫_{(x,y): 0 < x < y < z < 1}dxdydz = ∫₍₀₁₎∫₍ₓ₁₎∫_(y 1) dzdydx
                                                     = ∫₍₀₁₎∫₍ₓ₁₎(1-y)dydx
                                                     = ∫₍₀₁₎(x²/2 - x + 1/2)dx
                                                     = (1³/6 - 1²/2 + 1/2(1)) - 0
                                                     = 1/6
(b) If c is the subset {(x,y,z): 0 < x = y = z < 1} compute Q(c).
∫∫∫_{(x,y,z): 0 < x = y = z < 1}dxdydz = 0

1.2.13 let C denote the set {(x,y,z): x² + y² + z² <= 1. Using spherical coordinates, evaluate
Q(c) = ∫∫∫_c Sqrt(x² + y² + z²) dxdydz
We know x² + y² + z² = r² -> Sqrt(x² + y² + z²) = r
Given x = rsinφcosθ, y = rsinφsinKh*, and z = rcosφ we have dxdydz = r²sinθdrdφdθ
∫∫∫_c Sqrt(x² + y² + z²) dxdydz = ∫_(0,2π)∫_(0,π)∫_(0,1)r³sinθdrdφdθ
                                             = π

1.2.14 To join a certain club, a person must either be a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?
|A ∪ B| = |A| + |B| - |A ∩ B|
Let A be the set of statisticians and B be the set of mathematicians
Then |A ∪ B| = 25, |A| = 19, and |B| = 16
Hence 25 = 19 + 16 - |A ∪ B|
-> |A ∪ B| = 10
Therefore, 10 persons are both a statistician and a mathematician

1.2.15 After a hard-fought football game, it was reported that, of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, and 3 hurt both a hip and an arm, 2 hurt both a hip and knee, and 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.
|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|
Let A be the set of hurt hips, let B be the set of hurt arms, C be the set of hurt knees
11 = 8 + 6 + 5 - 3 - 2 - 1 - 0 = 13
There were only 11 players but according to the report there should be 13.

1.3 The Probability Set Function

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1.3.1 Definition and Basic Properties
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We are interested in assigning probabilities to evens, that is subsets of the sample space Ξ.

If Ξ is finite then we could take the set of all subsets as the collection.

We assume that in all cases the collection of events is sufficiently rich to include all possible events and is closed under complements and countable unions of these events.

Given the assumption above and DeMorgan's law we conclude that the collection is then also closed under countable intersections.

A collection withe properties stated above is known as a σ-field of subsets, we will denote this collection as Β.

Once a sample space Ξ and collection of events Β is exist one can define the third component of a probability space known as a Probability Set Function.

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Probability Set Function : :
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Definition 1.3.1 (Probability)
Let Ξ be a sample space and let Β be the set of events. Let P be a real-valued function defined on Β. Then P is a Probability Set Function if P satisfies the following three conditions:
(1) P(A) >= 0, for all A ∈ B
(2) P(Ξ) = 1
(3) If {Aₙ} is a sequence of events in B and Aₘ ∩ Aₙ = ∅ for all m /= n, then P(∪Aₙ) = ∑P(Aₙ)

(1) states that if some event A is in the collection B then then it has to have an assigned probability and it cannot be negative.

(2) Given the assumption that an event must happen and that the set of all events is Ξ, then the probability of any event happening, P(Ξ), must be equal to 1. That is we are absolutely certain that an event will occur.

(3) tells us that a collection of events whose members are pairwise disjoint is called a mutually exclusive collection and its union is referred to as a disjoint union. Pairwise joint sets are sets whose intersection results in ∅, hence they share no members. Given they share no members, computing the probability of a mutually exclusive union is equivalent to computing the probability of each disjoint set and then taking the union between them. Likewise, if a set of events are in a mutually exclusive collection then we can compute the probability of each event individually and then add all of those probabilities up to obtain the probability of the sequence of events.

A mutually exclusive and exhaustive collection of events forms a partition of Ξ.

A probability set function tells us how the probability is distributed over the set of events B.

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Properties of Probability Set Functions : :
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For all theorems we assume that P(A) is the probability set function defined on the collection of events B of a sample space Ξ.

Theorem 1.3.1
For each event A∈B, P(A) = 1 - P(Aᶜ)
Proof.
We know, Ξ = A ∪ Aᶜ and by (2) in definition 1.3.1 P(Ξ)=1
We know, A ∩ Aᶜ = ∅ and by (3) P(∪Aₙ) = ∑P(Aₙ)
It follows, 1 = P(A) + P(Aᶜ)
Which implies, P(A) = 1 - P(Aᶜ)

Theorem 1.3.2
The probability of the null set is zero; that is P(∅) = 0
Proof.
Let A = ∅
It follows, Aᶜ = Ξ
By (2) in 1.3.1 we get P(Ξ) = 1
By theorem 1.3.1 we get P(A) = 1 - P(Aᶜ)
P(A) = P(∅) = 1 - P(Aᶜ) = 1 - P(Ξ) = 1 - 1 = 0

Theorem 1.3.3
If A and B are events such A ⊂ B, then P(A) <= P(B).
Proof.
B = A U (Aᶜ ∩ B) and A ∩ (Aᶜ ∩ B) = ∅
Hence, by (3) of definition 1.3.1,
P(B) = P(A) + P(Aᶜ ∩ B)
From (1) of definition 1.3.1,
P(Aᶜ ∩ B) >= 0
Hence, P(B) >= P(A)

Theorem 1.3.4
For each A ∈ B, o <= P(A) <= 1
Proof.
Given ∅ ⊂ A ⊂ Ξ, by theorem 1.3.3 we have
P(∅) <= P(A) <= P(Ξ) or 0 <= P(A) <=1

Theorem 1.3.5
If A and B are events in Ξ, then P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Proof.
Each of the sets A ∪ B and B can be represented, respectively, as a union of non-intersecting sets as follows,
A ∪ B = A ∪ (Aᶜ ∩ B) and B = (A ∩ B) ∪ (Aᶜ ∩ B)
By (3) in definition 1.3.1,
P(A ∪ B) = P(A) + P(Aᶜ ∩ B) and P(B) = P(A ∩ B) + P(Aᶜ ∩ B)
It follows, P(Aᶜ ∩ B) = P(B) - P(A ∩ B)
Hence, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

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Equilikely : :
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Definition 1.3.2 (Equilikely Case)
Let Ξ = {x₁, x₂, . . . , xₘ} be a finite sample space. Let pᵢ = 1/m for all i=1,2,...,m and for all subsets A of Ξ define P(A) ∑1/m = #(A)/m where #(A) denotes the number of elements in A. Then P is a probability on Ξ and is referred to as the equilikely case.

Equilikely cases are common models for many encountered systems.

For instance, the flipping of a fair coin, drawing cards from a deck of 52, a spin of a spinner with numbers 1 to 36, are all examples of equilikely cases.

In such cases all we need to compute the probability of an event is the number of elements in an event.

1.3.1 Counting Rules

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Multiplication Rule : :
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Let A = {x₁,x₂,...,xₘ} be a set of m elements and let B = {y₁,y₂,...,yₙ} be a set of n elements.

There are mn ordered pairs (xᵢ,yⱼ), i=1,2,...,m and j=1,2,...,n, of elements.

Ex. Let there be five roads between cities I and II, and let there be 10 roads between cities II and III.
m*n = 5*10 = 50 ways of going from city I to city III.

This rule extends to more than two sets.

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Permutations : :
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Let A be a set with n elements.

Say we are interested in k-tuples whose components are elements of A.

By the extended mn rule there are n*n*...*n = nᵏ such k-tuples whose components are elements of A.

Suppose k <= n and we are interested in k-tuples whose components are distinct elements of A.

There are n elements from which to choose for the first component, n-1 fro the second component,
...,n-(k-1) for the kth.

Hence, by the mn rule there are n(n-1)*...*(n-(k-1)) such k-tuples with distinct tuples.

We call a k-tuple a Permutation and denote it with the symbol Pₖⁿ, that is the number of k permutations taken from a set of n elements.

Pₖⁿ = n(n-1)*...*(n-(k-1)) = n!/(n-k)!

Ex (Birthday Problem). Suppose there are n people in a room. Assume that n < 365 and that the people are unrelated in any way. Find the probability of event A that at least 2 people have the same birthday.
We assign 1 to n to the people in the room.
We use n-tuples to denote the birthdays of the first person through the nth person.
By mn rule there are 365ⁿ possible birthday n-tuples for these n people.
Hence, each n-tuple has a probability of 1/365ⁿ
The complement of A is the event that all of the birthdays are distinct
that is number of n-tuples fro Aᶜ is Pₙ³⁶⁵
Hence P(A) = 1 - Pₙ³⁶⁵/365ⁿ

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Combinations : :
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Let A be a set with n elements.

Order does not matter, we wish to count the subsets of k elements taken from A.

(ⁿₖ) denotes the total number of subsets.

Consider a subset of k elements from A, by the permutation rule it generates Pₖᵏ = k(k-1)...1 = k!permutations.

All of these permutations are distinct from the permutations generated be other subsets of k elements from A.

Each permutation of k distinct elements drawn from A must be generated by one of these subsets.

We have shown that Pₖⁿ = (ⁿₖ)k!
                        n!
Hence, (ⁿₖ) = ----------
                     k!(n-k)!

(ⁿₖ) is considered combinations of k things from a set of n things.

Ex (Poker Hands). Let a card be drawn at random from an ordinate deck of 52 playing cards that has been well shuffled.
The sample space Ξ consists of 52 elements, each element representing one and only one of the 52 cards. Because the deck is well shuffled we can assume that each outcome has a probability of 1/52.Let E₁ be the outcomes of spades, that is P(E₁) = 13/52 = 1/4
Let E₂ be the outcomes of that are kings, that is P(E₂) = 4/52 = 1/13
Now suppose we will draw five cards at random without replacement.
Order is not important so each hand is a subset of five elements drawn from a set of 52 elements.
Hence, (⁵²₅) hands.
if the deck is well shuffled each is equilikely hence 1/(⁵²₅).
We can now compute interesting hands from poker.
Let E₁ be the event of a flush, that is five cards of the same suite. There are (⁴₁) = 4 suits to choose for the flush, and for each suite there are (¹³₅) possible hands.

Hence,                                                            (⁴₁)(¹³₅)         4*1287
                                                          P(E₁) = ------------ = ------------- = 0.00198
                                                                          (⁵²₅)          2598960

1.3.2 Additional Properties of Probability

Consider the lim(n->∞)P(Cₙ)

Can we legitimately interchange the limit and P?

Theorem 1.3.6
Let {Cₙ} be a non-decreasing sequence of events. Then
lim(n->∞)P(Cₙ) = P(lim(n->∞)Cₙ) = P(∪Cₙ)
Let {Cₙ} be decreasing sequence of event events. Then
lim(n->∞)P(Cₙ) = P(lim(n->∞)Cₙ) = P(∩Cₙ)
Proof.
Define the sets, called rings, as R₁=C₁ for n>1, Rₙ = Cₙ ∩ Cᶜₙ₋₁.
It follows that ∪Cₙ = ∪Rₙ and that Rₘ ∩ Rₙ = ∅ for m ≠ n.
Also, P(Rₙ) = P(Cₙ) - P(Cₙ₋₁).
Applying the third axiom of probability,
P[lim(n->∞)Cₙ] = P(∪Cₙ) = P(∪Rₙ) = ∑P(Rₙ) = lim(n->∞)∑P(Rⱼ)
                                                                   = lim(n->∞){P(C₁) + ∑[P(Cⱼ) - P(Cⱼ₋₁)]}
                                                                   = lim(n->∞)P(Cₙ)

The above theorem shows that we can in fact interchange the limit and P.

Theorem 1.3.7 (Boole's Inequality)
Let {Cₙ} be an arbitrary sequence of events. Then P(∪Cₙ) <= ∑P(Cₙ)
Proof.
Let Dₙ = ∪Cᵢ.
Then {Dₙ} is an increasing sequence of events that go up to ∪Cₙ.
Also, for all j, Dⱼ = Dⱼ₋₁ ∪ Cⱼ.
Hence, by theorem 1.3.5
P(Dⱼ) <= P(Dⱼ₋₁) + P(Cⱼ)
that is,
P(Dⱼ) - P(Dⱼ₋₁) <= P(C)
In this case, the Cᵢs are replaced by the Dᵢs. Hence, using the inequality from theorem 1.3.6 and the fact that
P(C₁) = P(D₁) we have,
P(∪Cₙ) = P(∪Dₙ) = lim(n->∞){P(D₁) + ∑[P(Dⱼ) - P(Dⱼ₋₁)]} <= lim(n->∞)∑P(Cⱼ) = ∑P(Cₙ)

Boole's inequality states that for a sequence of events the probability that at least one of the events happens is no greater than the sum of the probabilities of the individual events.