Sa967St
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Yeah I have no idea. The wording is lol.
Mathematics marathon
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cmhardw
Premium Member
I know that what I am doing is not working, because I am getting a bogus answer, but here is what I did:
Bayes' Theorem approach
\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)
A = A 75 year old lives to 85
B = A 75 year old lives to 80
p(A|B) = probability that a 75 year old lives to 85 given that they lived to 80 (what the problem is asking)
p(B|A) = probability that a 75 year old lives to 80 given that they lived to 85 = 1
p(A) = 0.5
p(B) = unknown
\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)
\( p(A|B)=\frac{1*0.5}{p(B)} \)
\( p(A|B)=\frac{1}{2*p(B)} \)
Now examining p(B) we will examine how people live to 90.
A 75 year old has a 20% chance to live to 90.
an 80 year old has a 10% chance to live to 90.
Imagine there are N (N>0) 75 year olds in a group. 0.2N will survive to 90.
OR
some unknown portion, p(B), of them will survive to 80. Then, 10% of that group will live to 90.
So in an original group of 75 year olds, 0.2N of them live to 90 and N*p(B)*0.1 live to 90.
\( 0.2N = N*p(B)*0.1 \)
\( 2=p(B) \)
Which is clearly bogus. I'm thinking that I am either somehow not working the partition of people correctly, and am overcounting the number that survive, or otherwise I am not sure where I am going wrong.
That's my stab at it so far.
\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)
A = A 75 year old lives to 85
B = A 75 year old lives to 80
p(A|B) = probability that a 75 year old lives to 85 given that they lived to 80 (what the problem is asking)
p(B|A) = probability that a 75 year old lives to 80 given that they lived to 85 = 1
p(A) = 0.5
p(B) = unknown
\( p(A|B)=\frac{p(B|A)p(A)}{p(B)} \)
\( p(A|B)=\frac{1*0.5}{p(B)} \)
\( p(A|B)=\frac{1}{2*p(B)} \)
Now examining p(B) we will examine how people live to 90.
A 75 year old has a 20% chance to live to 90.
an 80 year old has a 10% chance to live to 90.
Imagine there are N (N>0) 75 year olds in a group. 0.2N will survive to 90.
OR
some unknown portion, p(B), of them will survive to 80. Then, 10% of that group will live to 90.
So in an original group of 75 year olds, 0.2N of them live to 90 and N*p(B)*0.1 live to 90.
\( 0.2N = N*p(B)*0.1 \)
\( 2=p(B) \)
Which is clearly bogus. I'm thinking that I am either somehow not working the partition of people correctly, and am overcounting the number that survive, or otherwise I am not sure where I am going wrong.
That's my stab at it so far.
Keroma12
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Not sure I'm right, but I did:
P(75-->90)=P(75-->85)xP(85-->90)
P(80-->90)=P(80-->85)xP(85-->90)
So P(75-->90)xP(80-->85)=P(75-->85)xP(80-->90)
So [.2]x[n]=[.5]x[.1]
So n=.25
P(75-->90)=P(75-->85)xP(85-->90)
P(80-->90)=P(80-->85)xP(85-->90)
So P(75-->90)xP(80-->85)=P(75-->85)xP(80-->90)
So [.2]x[n]=[.5]x[.1]
So n=.25
yamahammer08
Regular
This thread makes me feel stupid. I always used to be good at math, well at least I thought I was, till I hit college. I just stared at Integral (t+1)/t for a good 10 minutes before I realized what to do haha
cmhardw
Premium Member
Keroma,
I essentially did the same thing, but this result means that 200% of 75 year olds live to be 80, the same error I ran into.
Using your notation:
p(75-->80)*p(80-->85)*p(85-->90)=0.2
Now we need p(85-->90) which we will get from:
p(80-->90)=p(80-->85)*p(85-->90)
Rearranging we get:
p(80-->90) / p(80-->85) = p(85-->90)
Now that we have calculated that p(80-->85)=0.25 we can substitute:
0.1/0.25 = p(85-->90)
2/5 = p(85-->90)
Now going back to:
p(75-->80)*p(80-->85)*p(85-->90)=0.2
We will solve for p(75-->80)
p(75-->80) = 0.2 / [p(80-->85)*p(85-->90)]
p(75-->80) = 0.2 / [0.25 * 2/5]
p(75-->80) = 2
So 200% of all 75 year olds survive to 80. Or, the probability of being 75 and living to 80 is 200%. Take a look at my post for another take on the same approach, but I run into the same error.
I essentially did the same thing, but this result means that 200% of 75 year olds live to be 80, the same error I ran into.
Using your notation:
p(75-->80)*p(80-->85)*p(85-->90)=0.2
Now we need p(85-->90) which we will get from:
p(80-->90)=p(80-->85)*p(85-->90)
Rearranging we get:
p(80-->90) / p(80-->85) = p(85-->90)
Now that we have calculated that p(80-->85)=0.25 we can substitute:
0.1/0.25 = p(85-->90)
2/5 = p(85-->90)
Now going back to:
p(75-->80)*p(80-->85)*p(85-->90)=0.2
We will solve for p(75-->80)
p(75-->80) = 0.2 / [p(80-->85)*p(85-->90)]
p(75-->80) = 0.2 / [0.25 * 2/5]
p(75-->80) = 2
So 200% of all 75 year olds survive to 80. Or, the probability of being 75 and living to 80 is 200%. Take a look at my post for another take on the same approach, but I run into the same error.
Reading posts by Michael, Lucas, and Bruce sometimes makes me feel stupid as well. It's all relative. Math is very much about practice.
"An expert is a person who has made all the mistakes that can be made in a very narrow field." -Niels Bohr
The only way to make those mistakes is to practice it, and making the mistakes is the process of learning it.
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Rinfiyks
Regular
Rubik's cube inspired problem:
Imagine looking at a Rubik's cube from above - you'd just see the U face. But if you rotated the U face less than 90 degrees, you'd get something that looks like this:
What is the formula for the area of the black part of the image, with the angle the U face is rotated as a variable?
Imagine looking at a Rubik's cube from above - you'd just see the U face. But if you rotated the U face less than 90 degrees, you'd get something that looks like this:
What is the formula for the area of the black part of the image, with the angle the U face is rotated as a variable?
MaeLSTRoM
Veteran
Things we know
At all multiples of 90, the area is 0.
the area of the black section is 4x the area of each triangle, since each one would be identical.
The path traveled by the cube is a circle.
Calculating the area
Edit: Yeah i think this is wrong...
At all multiples of 90, the area is 0.
the area of the black section is 4x the area of each triangle, since each one would be identical.
The path traveled by the cube is a circle.
Calculating the area
We know that the area starts at (0,0) [angle,area]
It then goes to (45,max)
and then to (90,0)
I would then say that the area =
\( (sin(x).cos(x)) \) where x is the angle.
Thiis is true when the length of the cube is 1 becuase it would be 4 lost of the above formula, one for each triangle. Then you would have to divide by 4 becuase it would have to be less than 1 (the area never exceeds the area of the cube)
It then goes to (45,max)
and then to (90,0)
I would then say that the area =
\( (sin(x).cos(x)) \) where x is the angle.
Thiis is true when the length of the cube is 1 becuase it would be 4 lost of the above formula, one for each triangle. Then you would have to divide by 4 becuase it would have to be less than 1 (the area never exceeds the area of the cube)
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qqwref
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Interesting idea, Rinfiyks.
There are 4 black triangles, so we'll calculate the area of one of them and multiply by 4. Assume (for convenience) that the face is rotated clockwise by t degrees (t<90) and that the length of each side of the face/cube is 1 unit. I can't think of any really clever way to do this right now, so let's use a coordinate approach: center is (0,0), bottom right is (.5, -.5), and top right is (.5, .5).
Now, let's look at a triangle, say the bottom-right one. The triangle is defined by three lines: x=0.5, y=-0.5, and this line that goes between the top right corner rotated by t degrees, and the bottom right corner rotated by t degrees. So that third line goes through these two points:
(.5 cos(t) + .5 sin(t), -.5 sin(t) + .5 cos(t))
(.5 cos(t) - .5 sin(t), -.5 sin(t) - .5 cos(t))
The slope of this line (defined by (y2-y1)/(x2-x1)) is cot(t), and using the equation y = cot(t) x + b we can get that b = -.5 csc(t).
So our third line should be y = cot(t) x - .5 csc(t). It intersects the two other lines at the points (.5, .5 cot(t) - .5 csc(t)) and (-.5 tan(t) + .5 sec(t), -.5). This means that the height of the triangle is (.5 cot(t) - .5 csc(t) + .5) and the width is (.5 + .5 tan(t) - .5 sec(t)). The area of the triangle is (h*w)/2, or, in simplified form:
1/4 (1 - csc(t))(1 - sec(t)).
So the area of the entire black space would be 4 times that, or (1 - csc(t))(1 - sec(t)). A little icky, but it does have the appropriate property (from 0 to 90 degrees) of symmetry with the ends at 0. The maximum (at 45 degrees) evaluates to 3 - 2 sqrt(2), or about 0.172.
Now, let's look at a triangle, say the bottom-right one. The triangle is defined by three lines: x=0.5, y=-0.5, and this line that goes between the top right corner rotated by t degrees, and the bottom right corner rotated by t degrees. So that third line goes through these two points:
(.5 cos(t) + .5 sin(t), -.5 sin(t) + .5 cos(t))
(.5 cos(t) - .5 sin(t), -.5 sin(t) - .5 cos(t))
The slope of this line (defined by (y2-y1)/(x2-x1)) is cot(t), and using the equation y = cot(t) x + b we can get that b = -.5 csc(t).
So our third line should be y = cot(t) x - .5 csc(t). It intersects the two other lines at the points (.5, .5 cot(t) - .5 csc(t)) and (-.5 tan(t) + .5 sec(t), -.5). This means that the height of the triangle is (.5 cot(t) - .5 csc(t) + .5) and the width is (.5 + .5 tan(t) - .5 sec(t)). The area of the triangle is (h*w)/2, or, in simplified form:
1/4 (1 - csc(t))(1 - sec(t)).
So the area of the entire black space would be 4 times that, or (1 - csc(t))(1 - sec(t)). A little icky, but it does have the appropriate property (from 0 to 90 degrees) of symmetry with the ends at 0. The maximum (at 45 degrees) evaluates to 3 - 2 sqrt(2), or about 0.172.
Carrot
Veteran
so..
75->85=0.5
75->90=0.2
80->90=0.1
85->90=[75->90]/[75->85]=0.2/0.5=0.4
75->80=[75->90]/[80->90]=0.2/0.1=2 wtf?? (why are people cloning themself ??)
80->85=[75->85]/[75->80]=0.5/2=0.25
75->90=0.2
80->90=0.1
85->90=[75->90]/[75->85]=0.2/0.5=0.4
75->80=[75->90]/[80->90]=0.2/0.1=2 wtf?? (why are people cloning themself ??)
80->85=[75->85]/[75->80]=0.5/2=0.25
StachuK1992
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Gah. I've tried both of these, for so long now. I can't figure them out.
qqwref
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Not to spoil your fun, but you should probably stop trying:
WolframAlpha's solutions for both of these involve the Dilogarithm function, which strongly suggests that no solution with only simple functions (ln, trig, arc-trig) exists.
MaeLSTRoM
Veteran
Gah. I've tried both of these, for so long now. I can't figure them out.
The first is
\( -tan(x)(ln(sin(x))-1) \)
This is becuase \( \int ln(x) \) \( = x(ln(x)-1) \)
substitute in \( sin(x) \) and use inverse chain rule to get:
\( (sin(x)/-cos(x))(ln(sin(x)-1)) \) which then simplifies.
This is becuase \( \int ln(x) \) \( = x(ln(x)-1) \)
substitute in \( sin(x) \) and use inverse chain rule to get:
\( (sin(x)/-cos(x))(ln(sin(x)-1)) \) which then simplifies.
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Deluchie
Regular
t=-0.5x2 + 24x + 13, for x in terms of t
Keroma12
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Ok, well it says \( -\int \frac{ln(1-x)}{x} = Li2(x) \) so if we can split it up so that it contains a \( \frac{ln(1-x)}{x} \) as one term, it should be possible.
qqwref
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I don't think you're allowed to do this.
caseyd
Regular
you know, I used to like math
but this, this
ugh
but this, this
ugh
Rinfiyks
Regular
In general,
\( (a+b+c)^3 = (a^3 + b^3 + c^3) + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
I can't think of a clever way to go from here other than analyse each term and substitute in your head. I'll write a 0 if it's not an odd power of x. If it is, I'll write the coefficient. Starting from the left.
First bracket: 0 + 8 + 0 + 3(2 + 0 + 0 + 0 + 0 + 18) + 36
Second bracket: 0 - 8 + 0 + 3(- 2 + 0 + 0 + 0 + 0 - 18) + 36
So... (8 + 60 + 36) - (-8 - 60 + 36)
= 136
\( (a+b+c)^3 = (a^3 + b^3 + c^3) + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
I can't think of a clever way to go from here other than analyse each term and substitute in your head. I'll write a 0 if it's not an odd power of x. If it is, I'll write the coefficient. Starting from the left.
First bracket: 0 + 8 + 0 + 3(2 + 0 + 0 + 0 + 0 + 18) + 36
Second bracket: 0 - 8 + 0 + 3(- 2 + 0 + 0 + 0 + 0 - 18) + 36
So... (8 + 60 + 36) - (-8 - 60 + 36)
= 136
That looks right to me! I can't remember the answer exactly but that sounds familiar. I think I see where I went wrong in this now.
Rinfiyks
Regular
If you know how to expand (a+b)^3, you can substitute to expand (a+b+c)^3:
\( (a+b+c)^3 \)
Let \( b + c = d \)
\( (a+d)^3 \)
\( = a^3 + d^3 + 3(a^2d + d^2a) \)
Then substitute d out and simplify.
\( = a^3 + (b+c)^3 + 3(a^2(b+c) + (b+c)^2a \)
\( = a^3 + (b^3 + c^3 + 3(b^2c + c^2b)) + 3(a^2b+a^2c + (b^2+c^2 + 2bc)a) \)
\( = a^3 + b^3 + c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
Lots of places to make a mistake though (especially when typing it up, which I did, and took me a while to find it
)
Let \( b + c = d \)
\( (a+d)^3 \)
\( = a^3 + d^3 + 3(a^2d + d^2a) \)
Then substitute d out and simplify.
\( = a^3 + (b+c)^3 + 3(a^2(b+c) + (b+c)^2a \)
\( = a^3 + (b^3 + c^3 + 3(b^2c + c^2b)) + 3(a^2b+a^2c + (b^2+c^2 + 2bc)a) \)
\( = a^3 + b^3 + c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc \)
Lots of places to make a mistake though (especially when typing it up, which I did, and took me a while to find it
)
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