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The Math Help Thread
- Thread starter -COOLIO-
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SniperHunter
Banned
anyone?
Sorry, I can't say I have used that program.
BTW, anyone has found a solution to my problem? XD
ChainsOfFate
Neo Member
I have the following complexity function:
f(0) = 2m
f= 4f(n-1)^2+f(n-1)
Is it possible to find f in terms of m?
Is the question simply whether it is possible, or what the actual function would be for any given integer value of n?
The second.
SniperHunter
Banned
which program do you use to create graphs? fuck I need to hand this in soon, so many delays!!
ChainsOfFate
Neo Member
Damn. Any hopes I had of finding a nice pattern to the recursion were dashed once I got to f(3).
SniperHunter
Banned
come on guys whats the best way to draw a tangent line for non polynomial and non quadratic graph. the graph basically depicts an orange falling down.
I'm not sure if I understand the problem. Do you have a bunch of measurement data that you plot in a graph? If that is the case, I think the best way to get the tangent is to fit a polynomial curve to your data points and then calculate the tangent line from that.
Anticitizen One
Banned
So I just started Precalc after not taking a Math course in years and i'm really strugglign with the concept of radians. I'm stuck on a problem where you have to find the coterminal of 25(pi)/4 and i understand you basically keep adding/subtracting by 2(pi) to find it. However when I try to do this on my calculator (TI-83 plus) it keeps giving me answers in decimals and I can't convert them to fractions. When I look at example problems (like this http://hotmath.com/hotmath_help/topics/coterminal-angles.html) They use fractions all the way through but I have no idea how to do that.
For example, Example #2 on this page http://hotmath.com/hotmath_help/topics/coterminal-angles.html says find a positive and negative coterminal of (pi)/3 but I don't understand how adding 2(pi) to it becomes 7(pi)/3? Where does that 7 come from? It does match up when I do the division on my calculator but I have no clue how to do this using fractions. Can anyone help me out?
For example, Example #2 on this page http://hotmath.com/hotmath_help/topics/coterminal-angles.html says find a positive and negative coterminal of (pi)/3 but I don't understand how adding 2(pi) to it becomes 7(pi)/3? Where does that 7 come from? It does match up when I do the division on my calculator but I have no clue how to do this using fractions. Can anyone help me out?
hemtae
Member
well 2(pi) = 6(pi)/3 because 2 = (6/3). So when you do the addition its (1+6)(pi)/3 = 7(pi)/3.
ActStriker
Member
This is basic fraction math, actually.
Ignore PI for a second; it's basically saying this: 1/3 + 2 = 7/3, or in other words, 1/3 + 6/3 = 7/3.
If you go into set up (shift -> mode/set up on mine) you should be able to change it to RAD (radians) so everything appears like that. Then you can press S<=>D to switch back and forth with decimals. If you are doing much more maths you will probably use radians a lot as they are loads easier to deal with.
hemtae
Member
9 and -2
of course!!!!
Could someone help with a really basic problem? It's about graphing ellipses. Here's the standard form of an ellipse:
(x - h) / a^2 + (y - k) / b^2
If the a^2 is under the x, it's a horizontal ellipse, but it if it's under the y, it's vertical.
(x - h) / b^2 + (y - k) / a^2
So how do I know whether it's an a or a b if the problem is with numbers? I have a math problem telling me to graph the ellipse and the equation is:
Here's my answer.
Can some tell me if it's right? The main problem I have is that I don't know if it's vertical or horizontal.
(x - h) / a^2 + (y - k) / b^2
If the a^2 is under the x, it's a horizontal ellipse, but it if it's under the y, it's vertical.
(x - h) / b^2 + (y - k) / a^2
So how do I know whether it's an a or a b if the problem is with numbers? I have a math problem telling me to graph the ellipse and the equation is:
Here's my answer.
Can some tell me if it's right? The main problem I have is that I don't know if it's vertical or horizontal.
Could someone help with a really basic problem? It's about graphing ellipses. Here's the standard form of an ellipse:
(x - h) / a^2 + (y - k) / b^2
If the a^2 is under the x, it's a horizontal ellipse, but it if it's under the y, it's vertical.
(x - h) / b^2 + (y - k) / a^2
So how do I know whether it's an a or a b if the problem is with numbers? I have a math problem telling me to graph the ellipse and the equation is:
Here's my answer.
Can some tell me if it's right? The main problem I have is that I don't know if it's vertical or horizontal.
If I remember correctly, a squared is always the larger number, so in this case, a^2 = 16 (which means that a^2 is dividing y), so that means the ellipse's major axis is vertical.
Treefingers
Member
The bolded is true only under the assumption that a > b . So by that definition a^2 is always the larger of the two denominators.Could someone help with a really basic problem? It's about graphing ellipses. Here's the standard form of an ellipse:
(x - h) / a^2 + (y - k) / b^2
If the a^2 is under the x, it's a horizontal ellipse, but it if it's under the y, it's vertical.
(x - h) / b^2 + (y - k) / a^2
So how do I know whether it's an a or a b if the problem is with numbers? I have a math problem telling me to graph the ellipse and the equation is:
Here's my answer.
Can some tell me if it's right? The main problem I have is that I don't know if it's vertical or horizontal.
My rule of thumb is, if the denominator under the y term is larger than the one under the x term, then it's more stretched out in the y direction than in the x direction, and vice versa. So in this case 16 > 4 so it's wider in the y direction. And a would be the larger number so a = 4 and b = 2.
Joel Was Right
If you smell burning, it's probably the generators acting up. Report it anyway.
nm
youngwerther
Banned
I'm at the very beginning of my Calculus 1 class and want to make sure I'm doing this right.
I'm calculating slopes of secant lines for various x values.
My given point P is (0.5,0)
Q is (x, cos pi x)
First x value was 0. That one was easy, got slope -2.
Second x value was 0.4. Got slope -9.997595.
Third x value was 0.49. Got slope -99.9639097. Errr...ok..
Does this look like it's going well to anyone? I ask because I know they are trying to convey the concept of limits, but this being my first week in a Calculus course, I've never seen the slopes go like that. I mean I've seen like 1.11111 to 1.010101 to 1.00101010 to 1.00010101, but for it to jump from -9 to -99 threw me off and makes me think I'm doing something wrong.
Probably a really dumb question but I have to do like 10 variables for every fucking question and if I'm doing it wrong I want to find out sooner rather than later.
I'm calculating slopes of secant lines for various x values.
My given point P is (0.5,0)
Q is (x, cos pi x)
First x value was 0. That one was easy, got slope -2.
Second x value was 0.4. Got slope -9.997595.
Third x value was 0.49. Got slope -99.9639097. Errr...ok..
Does this look like it's going well to anyone? I ask because I know they are trying to convey the concept of limits, but this being my first week in a Calculus course, I've never seen the slopes go like that. I mean I've seen like 1.11111 to 1.010101 to 1.00101010 to 1.00010101, but for it to jump from -9 to -99 threw me off and makes me think I'm doing something wrong.
Probably a really dumb question but I have to do like 10 variables for every fucking question and if I'm doing it wrong I want to find out sooner rather than later.
Treefingers
Member
I think you're on degree mode on your calculator.
Make sure you're using radians.youngwerther
Banned
oh,
wow.
Thanks dude.
wow.
Thanks dude.
While your problem was definitely caused by using degrees instead of radians, that does not make your original calculations invalid, it is reasonable for the slope to jump this rapidly in certain situations. Think about what the slope would be between two points with different y value but identical x.
Could someone help with a really basic problem? It's about graphing ellipses. Here's the standard form of an ellipse:
(x - h) / a^2 + (y - k) / b^2
If the a^2 is under the x, it's a horizontal ellipse, but it if it's under the y, it's vertical.
(x - h) / b^2 + (y - k) / a^2
So how do I know whether it's an a or a b if the problem is with numbers? I have a math problem telling me to graph the ellipse and the equation is:
Here's my answer.
Can some tell me if it's right? The main problem I have is that I don't know if it's vertical or horizontal.
Just set x=5 first, then solve for y, this gives y+3=+-4, so y=1 and y=-7 are the vertical maxima. Then set y=-3 and solve for x, giving x-5=+-2, therefore x=7 and x=3 are the horizontal maxima. Then (deltax=4)<(deltay)=8, so it's streched in the vertical y direction. In short: yes, that graph is correct.
youngwerther
Banned
I need more help with this Calculus homework. I kind of feel like I've been thrown to the wolves here (the wolves being James Stewart).
If a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y=40t - 16t^2.
Part A. Find the average velocity for the time period beginning when t=2 and lasting
(i) 0.5 seconds.
(ii) 0.1 seconds.
(iii) 0.05 seconds.
(iv) 0.01 seconds.
_____________________________________________________________________
Naturally there's a part B (which I'll probably need help with later on) but for now I just want to understand part A.
Ok, so, find the average velocity. Distance traveled divided by time taken.
Well, it says my equation gives me height in feet, so there is my distance traveled.
Now, for my time taken. Here's where things get hazy.
Are they saying that the time taken is .5, .1, et cetera?
Or are they saying that the time taken is 2 MINUS those values?
Even more confusing is in the back of the book, the answer for all the variables in part A is a negative number.
Specifically
(i) -32ft/s
(ii) - 25.6 ft/s
(iii) -24.8 ft/s
(iv) -24.16 ft/s
So either way I'm still lost, because I don't know how they are getting any negative number.
If a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y=40t - 16t^2.
Part A. Find the average velocity for the time period beginning when t=2 and lasting
(i) 0.5 seconds.
(ii) 0.1 seconds.
(iii) 0.05 seconds.
(iv) 0.01 seconds.
_____________________________________________________________________
Naturally there's a part B (which I'll probably need help with later on) but for now I just want to understand part A.
Ok, so, find the average velocity. Distance traveled divided by time taken.
Well, it says my equation gives me height in feet, so there is my distance traveled.
Now, for my time taken. Here's where things get hazy.
Are they saying that the time taken is .5, .1, et cetera?
Or are they saying that the time taken is 2 MINUS those values?
Even more confusing is in the back of the book, the answer for all the variables in part A is a negative number.
Specifically
(i) -32ft/s
(ii) - 25.6 ft/s
(iii) -24.8 ft/s
(iv) -24.16 ft/s
So either way I'm still lost, because I don't know how they are getting any negative number.
They want you to use y(2) as a start point and then use y(2+x) as the end point. Velocity is defined as the end point minus the start point divided by the time taken. Remember also that velocity is not the same as speed. Speed is just the magnitude, velocity is magnitude and direction hence why it can be negative.
youngwerther
Banned
Hrmph.
Ok, so it goes like this?
(i)
Start point t=2
End point t=(2+0.5) -> t=2.5
height at start point
y=40(2) - 16(2^2) = 16
height at end point
y=40(2.5) - 16(2.5^2) = 0
Average velocity -
(0 - 16)/0.5 = -32
And I suppose that the answer is negative means that it is either losing speed, or is traveling in the opposite direction that it was originally thrown due to gravity?
The height at the end point being 0 kind of threw my off. Wouldn't that mean the ball is sitting on the ground?
That seems to work I guess.
Thanks, I would have never thought to use the equation like that. I feel really stupid when I try to do this homework, and it's only the first week.
I'm sure I'll be back in about 30 seconds for part B.
Ok, so it goes like this?
(i)
Start point t=2
End point t=(2+0.5) -> t=2.5
height at start point
y=40(2) - 16(2^2) = 16
height at end point
y=40(2.5) - 16(2.5^2) = 0
Average velocity -
(0 - 16)/0.5 = -32
And I suppose that the answer is negative means that it is either losing speed, or is traveling in the opposite direction that it was originally thrown due to gravity?
The height at the end point being 0 kind of threw my off. Wouldn't that mean the ball is sitting on the ground?
That seems to work I guess.
Thanks, I would have never thought to use the equation like that. I feel really stupid when I try to do this homework, and it's only the first week.
I'm sure I'll be back in about 30 seconds for part B.
The negative velocity does indeed mean that it is going in the opposite direction. And yes, if the height is zero that does mean that it is lying on the ground (unless zero is defined at another point, technically you can set zero wherever you want).
The Awesomest
Member
Hi MathGAF, I'm a history major who likes solving math problems now and then for the heck of it. My friend posed the following problem to me and I can't figure it out:
Now my initial guess was 105mph, because the first half was driven at half of 70, so I'd drive the second half at 70 + the missing half of 70, which equals 105. And (105 + 35)/2 = 70. But apparently I'm wrong. He says that 290 is relevant.
What am I missing? How should I set this problem up?
Now my initial guess was 105mph, because the first half was driven at half of 70, so I'd drive the second half at 70 + the missing half of 70, which equals 105. And (105 + 35)/2 = 70. But apparently I'm wrong. He says that 290 is relevant.
What am I missing? How should I set this problem up?
cpp_is_king
Member
Just a warning. I'm going to leave everything in fraction form the entire time and not calculate anything out with decimals, since lots of things cancel otu and it's easier for me this way than having to work with approximate decimals.
For the average speed to be 70mph you need the entire trip to take 290/70 hours.
You've already gone 105/2 miles at 35 mph, so you've spent a total of 105/70 hours on your trip when you reach the midpoint. Thus you have 290/70 - 105/70 = 185/70 hours remaining to complete the remaining 105/2 miles. In order to do this, your speed should be (105/2) / (185/70) = (105*70)/(185*2) = 19.86 mph
Edit: I'm stupid because I somehow used 2 different numbers for the total distance of the trip
In any case, the above is the reasoning you would use for how to solve the problem, assuming you put the numbers in correctly. But as rjc points out, it appears to be impossible
JustProgress
Member
I've seen problems like this before. It's probably impossible to do so. Sounds like an IB interview problem.
fake e: nvm
I guess by "first half of the trip" he means the first 145 miles. So if you travel the first 145 miles at 35 mph it will take you 145/35 = 4.142857... hours. But to average 70 mph for the whole trip, the whole trip would have to take 290/70 = 4.142857... hours. So you can't average 70 mph for the whole trip unless you figure out a way to travel infinitely fast for the second half.
The Awesomest
Member
OK thanks guys. What a jerk.
cpp_is_king
Member
I guess he could also have meant "first half of the trip" as in time. In which case it actually is possible. First half of the trip took 290/140 hours, from there compute the distance he travelled in that amount of time, subtract from 290 to find the remaining distance, then divide that distance by 290/140 again to find out how fast he has to go during the remainder of the trip. When I do this I get about 131.8 mph, but knowing me I probably did something stupid again
The total time required would be 290/70=4.14...h. Half of that is 4.14.../2=2.07...h which is how long you traveled at 35 mph. This gives a distance of 35*2.07...=72.5 miles. 290-72.5=217.5 miles, which is how long you must travel in the remaining 2.07... h giving a speed of 217.5/2.07...=105 mph. So based on this assumption the 105 mph answer is the correct one (unless I also made some stupid mistake).
Hypno Funk
Member
Your assumption is right if you assume that half of the journey TIME is done at 35 mph. If it is assumed that half of the journey DISTANCE is done at 35 mph, then as was pointed out this can't be done.
cpp_is_king
Member
Yea you'er right. Fucking numbers, math is so much easier when all you have are variables, lol.
Can anybody help me with this rate of convergence problem?
lim sin (1/(n^2)) = 0
as n -> infinite
I am having trouble with the textbooks explanation/example.
edit: This is my process for what I think is the answer. |sin (1/(n^2)) - 0 (because the lim equals 0)| = sin (1/(n^2)) <= 1/(n^2). Therefore 1/(n^2) is an upper bound, therefore the rate of convergence is O(1/(n^2)).
lim sin (1/(n^2)) = 0
as n -> infinite
I am having trouble with the textbooks explanation/example.
edit: This is my process for what I think is the answer. |sin (1/(n^2)) - 0 (because the lim equals 0)| = sin (1/(n^2)) <= 1/(n^2). Therefore 1/(n^2) is an upper bound, therefore the rate of convergence is O(1/(n^2)).
youngwerther
Banned
If x is -pi/2
then 1 + sinx is 0, right?
then 1 + sinx is 0, right?
When doing problems like this use theorems you know about convergence. I don't know what you've covered but I would try the Theorem that states that:
If (Kn) is a sequence that converges to 0, and (An) converges to a number a, then the sequence (An) converges to a with a rate of convergence O(Kn) if there exists a positive number R such that
lim |An - a| / |Kn| = R
as n -> infinite
So for your case, we can use two different methods. The easier method is that we know
lim sin(x) / x = 1
as x -> 0
So we substitute x for 1/n^2. Then we get
lim sin(1/n^2) / (1/n^2) = 1
as n -> infinity
(Note that now it's n -> infinity because when x = 1/n^2 and x goes to 0, then n goes to infinity)
This now satisfies the theorem above. sin(1/n^2) does converge to a number, a = 0, 1/n^2 does converge to 0, and R = 1, which is a positive constant.
This means according to
lim |An - a| / |Kn| = R
as n -> infinite
The rate of convergence is O(Kn) so for us it is O(1/n^2)
Done.
The other way would be to do a Taylor expansion of Sin(x) and use the same theorem above.
you're missing an exponent in the first line there, I'm going to assume though that it was:
12(w^4) + 26(w^3) + 12(w^2)
Pulling out 2(w^2) looks like a good step.
2(w^2) [6(w^2) + 13w + 6]
Then factor. I think you also had the above equation, but you didn't factor correctly. A good way to figure out if your factoring is correct is to multiply the quantities back and check that it results in your original equation. In yours, (6w^2 - 9w) (-4w +6) = -24(w^3) + 72(w^2) - 54w. Again, all this is done assuming that the exponent for the first term in the original problem is 4. If that's not right tell me. So what you'd actually get is:
2(w^2) [(2w + 3)(3w + 2)]
Once again, a good way to check this is correct is to multiply with FOIL. Let's take a detour and check that quick:
(2w + 3)(3w + 2)=(2w)(3w) + (2w)(2) + (3)(3w) + (3)(2)=6(w^2) + 4w + 9w + 6=6(w^2) + 13w + 6
Cool! We factored right. And from here you should be done. Your final answer:
2(w^2) [(2w + 3)(3w + 2)]
Hope I guessed at the right exponent. If not: oh well.
Guys I need help with a Probability question because I don't understand the notation. "Suppose we roll two dice and let X and Y be the two numbers that appear. Find the distribution of |X-Y|".
What do the | bars mean? Do they mean "probability of"?
The answer is apparently 0:6/36 but I have no idea how they got that because I don't see any indication of needing to use a ratio.
What do the | bars mean? Do they mean "probability of"?
The answer is apparently 0:6/36 but I have no idea how they got that because I don't see any indication of needing to use a ratio.
cpp_is_king
Member
absolute value?
Well in the textbook it says that |A| denotes the number of points in the set A.
Problem is I have no idea what that means or how to tie it into my problem.
They give an example in the textbook that says:
"A survey of 1000 students revealed that 750 owned stereos, 450 owned cars, and 350 owned both. How many owned either a car or a stereo?
Solution:
|SUC| = |S| + |C| - |SnC| = 750 + 450 - 350 = 850"
Don't know how to relate that to my problem though since I have no numbers and I don't know what |X-Y| is.
Hypno Funk
Member
The "|" / Pipe Symbol or Modulus signs just denote the absolute value as previously said.
So, |5-3| = |3-5| = 2.
I would imagine that the question is asking for the distribution of different possible results.
You have two die, when you roll both you have a one in six (1/6) chance of rolling any one particular number, so to roll two 6's would be (1/6*1/6 = 1/36).
So as an example, let us find the distribution of |X-Y| = 0 (i.e. the two numbers we roll subtract to make 0).
E.g. Let 'a' be the number of the first die, and 'b' the number of the second die.
(a,b) such that ||
|a-b| = 0 || Probability(a,b)
__________________________________
(1,1) = |1-1| = 0 || (1/6)*(1/6) = 1/36
(2,2) = |2-2| = 0 || (1/6)*(1/6) = 1/36
(3,3) = |3-3| = 0 || (1/6)*(1/6) = 1/36
(4,4) = |4-4| = 0 || (1/6)*(1/6) = 1/36
(5,5) = |5-5| = 0 || (1/6)*(1/6) = 1/36
(6,6) = |6-6| = 0 || (1/6)*(1/6) = 1/36
Hence, the distribution of 0 (i.e. all the values of (a,b) such that |a-b| = 0) is the sum of the above probabilities which is (6*(1/36)) = 6/36 = 1/6.
Thus the distribution of 0 is 1/6 as stated in your book. If you tried to do this for all values of (a,b) such that |a-b| = 1 (i.e. the distribution of 1) you would find that
there are multiple values which satisfy this e.g. |4-3| = |3-4| = |5-4| = |6-5| = 1 to name a few. Hence you would expect the distribution of |a-b| = 1 to be much larger than that of 0 (1/6).
Hope that helps.
Hypno Funk
Member
To continue from my previous post: the distribution of 1 would be as follows:
(a,b) such that ||
|a-b| = 1 || Probability(a,b)
__________________________________
(1,2) = |1-2| = 1 || (1/6)*(1/6) = 1/36
(2,1) = |2-1| = 1 || (1/6)*(1/6) = 1/36
(2,3) = |2-3| = 1 || (1/6)*(1/6) = 1/36
(3,2) = |3-2| = 1 || (1/6)*(1/6) = 1/36
(3,4) = |3-4| = 1 || (1/6)*(1/6) = 1/36
(4,3) = |4-3| = 1 || (1/6)*(1/6) = 1/36
(4,5) = |4-5| = 1 || (1/6)*(1/6) = 1/36
(5,4) = |5-4| = 1 || (1/6)*(1/6) = 1/36
(5,6) = |5-6| = 1 || (1/6)*(1/6) = 1/36
(6,5) = |6-5| = 1 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 1 is (10*(1/36)) = 10/36
cont.
(a,b) such that ||
|a-b| = 2 || Probability(a,b)
__________________________________
(1,3) = |1-3| = 2 || (1/6)*(1/6) = 1/36
(3,1) = |3-1| = 2 || (1/6)*(1/6) = 1/36
(2,4) = |2-4| = 2 || (1/6)*(1/6) = 1/36
(4,2) = |4-2| = 2 || (1/6)*(1/6) = 1/36
(3,5) = |3-5| = 2 || (1/6)*(1/6) = 1/36
(5,3) = |5-3| = 2 || (1/6)*(1/6) = 1/36
(4,6) = |4-6| = 2 || (1/6)*(1/6) = 1/36
(6,4) = |6-4| = 2 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 2 is (8*(1/36)) = 8/36 = 2/9
(a,b) such that ||
|a-b| = 3 || Probability(a,b)
__________________________________
(1,4) = |1-4| = 3 || (1/6)*(1/6) = 1/36
(4,1) = |4-1| = 3 || (1/6)*(1/6) = 1/36
(2,5) = |2-5| = 3 || (1/6)*(1/6) = 1/36
(5,2) = |5-2| = 3 || (1/6)*(1/6) = 1/36
(3,6) = |3-6| = 3 || (1/6)*(1/6) = 1/36
(6,3) = |6-3| = 3 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 3 is (6*(1/36)) = 6/36 = 1/6
(a,b) such that ||
|a-b| = 4 || Probability(a,b)
__________________________________
(1,5) = |1-4| = 4 || (1/6)*(1/6) = 1/36
(5,1) = |4-1| = 4 || (1/6)*(1/6) = 1/36
(2,6) = |2-5| = 4 || (1/6)*(1/6) = 1/36
(6,2) = |5-2| = 4 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 4 is (4*(1/36)) = 4/36 = 1/9
(a,b) such that ||
|a-b| = 5 || Probability(a,b)
__________________________________
(1,6) = |1-4| = 4 || (1/6)*(1/6) = 1/36
(6,1) = |4-1| = 4 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 5 is (2*(1/36)) = 2/36 = 1/18
So in total the distributions are:
(X,Y) such that |X-Y| = 0 : 6/36
(X,Y) such that |X-Y| = 1: 10/36
(X,Y) such that |X-Y| = 2: 8/36
(X,Y) such that |X-Y| = 3: 6/36
(X,Y) such that |X-Y| = 4: 4/36
(X,Y) such that |X-Y| = 5: 2/36
Notice that (6/36 + 10/36 + 8/36 + 6/36 + 4/36 + 2/36) = 36/36 = 1. Hence we have found the complete distribution. As we have found the complete list of possibilities for rolling two die.
(a,b) such that ||
|a-b| = 1 || Probability(a,b)
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(1,2) = |1-2| = 1 || (1/6)*(1/6) = 1/36
(2,1) = |2-1| = 1 || (1/6)*(1/6) = 1/36
(2,3) = |2-3| = 1 || (1/6)*(1/6) = 1/36
(3,2) = |3-2| = 1 || (1/6)*(1/6) = 1/36
(3,4) = |3-4| = 1 || (1/6)*(1/6) = 1/36
(4,3) = |4-3| = 1 || (1/6)*(1/6) = 1/36
(4,5) = |4-5| = 1 || (1/6)*(1/6) = 1/36
(5,4) = |5-4| = 1 || (1/6)*(1/6) = 1/36
(5,6) = |5-6| = 1 || (1/6)*(1/6) = 1/36
(6,5) = |6-5| = 1 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 1 is (10*(1/36)) = 10/36
cont.
(a,b) such that ||
|a-b| = 2 || Probability(a,b)
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(1,3) = |1-3| = 2 || (1/6)*(1/6) = 1/36
(3,1) = |3-1| = 2 || (1/6)*(1/6) = 1/36
(2,4) = |2-4| = 2 || (1/6)*(1/6) = 1/36
(4,2) = |4-2| = 2 || (1/6)*(1/6) = 1/36
(3,5) = |3-5| = 2 || (1/6)*(1/6) = 1/36
(5,3) = |5-3| = 2 || (1/6)*(1/6) = 1/36
(4,6) = |4-6| = 2 || (1/6)*(1/6) = 1/36
(6,4) = |6-4| = 2 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 2 is (8*(1/36)) = 8/36 = 2/9
(a,b) such that ||
|a-b| = 3 || Probability(a,b)
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(1,4) = |1-4| = 3 || (1/6)*(1/6) = 1/36
(4,1) = |4-1| = 3 || (1/6)*(1/6) = 1/36
(2,5) = |2-5| = 3 || (1/6)*(1/6) = 1/36
(5,2) = |5-2| = 3 || (1/6)*(1/6) = 1/36
(3,6) = |3-6| = 3 || (1/6)*(1/6) = 1/36
(6,3) = |6-3| = 3 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 3 is (6*(1/36)) = 6/36 = 1/6
(a,b) such that ||
|a-b| = 4 || Probability(a,b)
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(1,5) = |1-4| = 4 || (1/6)*(1/6) = 1/36
(5,1) = |4-1| = 4 || (1/6)*(1/6) = 1/36
(2,6) = |2-5| = 4 || (1/6)*(1/6) = 1/36
(6,2) = |5-2| = 4 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 4 is (4*(1/36)) = 4/36 = 1/9
(a,b) such that ||
|a-b| = 5 || Probability(a,b)
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(1,6) = |1-4| = 4 || (1/6)*(1/6) = 1/36
(6,1) = |4-1| = 4 || (1/6)*(1/6) = 1/36
So the distribution of |X-Y| = 5 is (2*(1/36)) = 2/36 = 1/18
So in total the distributions are:
(X,Y) such that |X-Y| = 0 : 6/36
(X,Y) such that |X-Y| = 1: 10/36
(X,Y) such that |X-Y| = 2: 8/36
(X,Y) such that |X-Y| = 3: 6/36
(X,Y) such that |X-Y| = 4: 4/36
(X,Y) such that |X-Y| = 5: 2/36
Notice that (6/36 + 10/36 + 8/36 + 6/36 + 4/36 + 2/36) = 36/36 = 1. Hence we have found the complete distribution. As we have found the complete list of possibilities for rolling two die.
Thank you so much! So "distribution" means to show the probabilities associated with each desired outcome, and then sum up all of the probabilities of the desired outcomes, more or less? Because 1/6 is the sum of the probabilities of each combination that gets you |X-Y| = 0.
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