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**mikau**- Replies: 2

This problem is driving me crazy. I got part(a) and I really think I'm doing (b) correctly but I'm not getting the same answer as my book.

**Componants of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defective components, 30% contain one defective component and 20% contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 and 2 defective componants being in the batch under each of the following conditions:**

**(a) Neither tested component is defective(b) One of the two tested components is defective**

Here's how i solved (a)

Let Bk be the event that k defective components are in a batch, and let Tk be the event that k of the two tested components are defective.

We seek P(B0 | T0), P(B1 | T0), and P(B2 | T0)

First, observing that Bk are mutually exclusive and exhaustive, we may use Bayes' theorem and compute these as:

P(Bi | T0)=P(T0 | Bi) P(Bi) /[ P(T0 | B0)P(B0) + P(T0 | B1)P(B1) + P(T0 | B2)P(B2)]

look at P(T0 | B0), P(T0 | B1) and P(T0 | B2).

P(T0 | B0) = 1, for if our batch contains no defective components, the probability of testing 0 defective components is 1.

P(T0 | B1) = (9 choose 2)/(10 choose 2), because we may test 2 out of all but the single defective component, and there are 10 choose 2 ways to test them total.

P(T0 | B2) = (8 choose 2)(10 choose 2), because we may test 2 out of all but the two defective components

Using these results in the formula for Bayes' theorem, gives me the correct answers for (a).

Now for (b), We seek P(B0 | T1), P(B1 | T1), and P(B2 | T1).

Using the same method as before, we now have to find the probabilities of P(T1 | B0), P(T1 | B1) and P(T1 | B2)

P(T1 | B0) = 0, for if there are no defective components, we cannot test 1 defective one

P(T1 | B1) = (9 choose 1)/(10 choose 2), as there are (9 choose 1) ways to select the remaining non-defective component to be tested out of (10 choose 2) ways total

P(T1 | B2) = (8 choose 1)/(10 choose 2), as there are (8 choose 1) ways to select the remaining non-defective component.

Plugging all this in to bayes' theorem gives me:

P(B0 | T1) = 0

P(B1 | T1) = 0.628

P(B2 | T1) = 0.372

Seems absolutely sound to me. However, my book claims the answers are:

0

0.457

0.543

I am particularly frustrated since my reasoning seemed to work for (a)

that makes sense.

So what kind of useful invarient are we looking for in the context of this topic?

no i'm allowing powers to be rational (right?)

so I think that doesn't include transcendental numbers.

I find this especially cool because it links linear algebra to number theory.

For instance! The fact that every number has a unique factorization follows from the fact that the primes form a basis of this vector space, and every element of a vector space can be expressed as a unique combination of its basis vectors.

Moreover, the scope of this goes beyond whole numbers because it includes numbers of the form (prime)^(rational).

So if I'm not mistaken, we can also use this to show that sqrt(2) =2^(1/2) is irrational. Because any fractional representation would be an attempt to create two distinct descriptions of sqrt(2) in terms of the basis vectors.

(edit) I take that back. I'm not sure you can prove the linear independence or primes without the FTA. Still I think the vector space notion is cool!

**mikau**- Replies: 10

I came up with this idea last night while brushing my teeth!

Let K denote the field of rationals.

Let U := { u | u = a[sub]1[/sub][sup]r1[/sup]a[sub]2[/sub][sup]r2[/sup]
a[sub]n[/sub][sup]rn[/sup]} where {a[sub]1[/sub]
a[sub]n[/sub]} are prime, and {r[sub]1[/sub]
r[sub]n[/sub]} are in K

For all u,v in U and k in K:

let u+v := uv,

let ku := u[sup]k[/sup],

let 0 in U := 1

Then we have U is a vector space over K.

**Proof:**

*Associativity: *

(u+v)+w = (uv)w = u(vw) = u+(v+w)

*Zero vector existence and properties:*

0+u = (1)u = u = u(1) = u+0

*Negative vector existence and properties:*

For every vector u = a[sub]1[/sub][sup]r1[/sup]a[sub]2[/sub][sup]r2[/sup]
a[sub]n[/sub][sup]rn[/sup], there exists (-u) = a[sub]1[/sub][sup]-r1[/sup]a[sub]2[/sub][sup]-r2[/sup]
a[sub]n[/sub][sup]-rn[/sup] such that:

u+(-u) = (-u)+u = (u)(u[sup]-1[/sup]) = 1 = 0

*Commutativity:*

u+v = uv = vu = v+u

*Distributivity of scalers:*

k(u+v) = k(uv) = (uv)[sup]k[/sup] = u[sup]k[/sup]v[sup]k[/sup]= (u[sup]k[/sup])+(v[sup]k[/sup]) = ku + kv

*Distributivity of vectors:*

u(k1+k2)= u[sup](k1+k2)[/sup]=u[sup]k1[/sup]u[sup]k2[/sup]= u[sup]k1[/sup]+u[sup]k2[/sup]=k1u + k2u

*Scaler Associativity:*

(ab)u = u[sup](ab)[/sup]=u[sup](ba)[/sup]= (u[sup]b[/sup])[sup]a[/sup]=a(u[sup]b[/sup])=a(bu)

*Unit Scaler:*

(1)u = u[sup]1[/sup]=u

**mikau**- Replies: 1

I found this a while ago but I think I forgot to show the good folks at mathisfun.

I was totally blown away by this. Its an implict equation that draws a Star Wars Tie Fighter!

http://www.leweyg.com/download/Screens/ImpView_StarWars.jpeg

you said hello 3 times! ;]

Cordic method huh? I'll look it up

That works too! :]

Originally, I thought of it this way: start with the point P= (1,0) and reflect it about the line L through the origin and Q= (1,x/2^n), to produce a point R

now set P equal to Q, and Q equal to R, and repeat. In a way its kind of like rolling a cone of length 1 and circumference x/2^n, through 2^n revolutions until the large end falls onto the point (sin(x),cos(x))

Nice! And of course we can say a[0]=x/2^M for the general case! B)

(edit) also i think it should say that a[M]=sin(x) and b[M]=cos(x), otherwise the above difference equation just describes a sequence, and we don't know where to stop it.

**mikau**- Replies: 7

Here is an amusing method for finding the sine of an angle that makes intuitive sense:

suppose we want to find the sine of 1 radian, about 57 degrees.

First we choose a power of 2 that is large relative to the angle. I'll choose 32.

1/32 is fairly small and so sin(1/32) ≈ 1/32, and cos(1/32) ≈ 1.

Now that we know sin(1/32) and cos(1/32) we can use 5 applications of the double angle identities: sin(2x) = 2sin(x)cos(x) and cos(2x)=cos^2(x) - sin^2(x) to get the final sine and cosine!

I worte a simple program to do it for me. Have a look:

sin(1/32): 0.03125 cos: 1

sin(2/32): 0.0625 cos: 0.999023

sin(4/32): 0.124878 cos: 0.994142

sin(8/32): 0.248293 cos: 0.972723

sin(16/32): 0.48304 cos: 0.884541

final (32/32): sin: 0.854537 cos: 0.549085

results by windows calculator:

sin: 0.8414709848078965066525023216303, cos: 0.54030230586813971740093660744298

Hey! Thats pretty close! Of course! you can make it more accurate by choosing a larger power of 2. Choosing 2^10 for instance yields: sin: 0.841881 cos: 0.540566

Its hardly the fastest way to do it, but its easy to understand why it works and I think that makes it fun!

Originally I envisioned this as a series of line reflections. But I eventually realized I was essentially using the double angle formulas. Maybe I'll post a diagram later.

thanks, Tigeree! :]

I don't see how Ax^2 + Bxy + Cy^2 can be turned into b^2-4ac.

how about B^2 - 4AC?

Its easy! Watch, if we have 5x^2 + 3xy + 2y^2 and we want to find the discriminant, we observe A=5, B=3, and C=2, so to find B^2 -4*AC we just substitute:

(3)^2 - 4*(5)*(2)= 9-40=-31.

Here's the generlal conic equation: Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

Basically you make different shapes by assigning values to the coefficients A,B,C,D, E and F. Usually what you do is make some of those values 0 so the terms dissapear! For instance,

let A=1,B=0,C=1,D=0, E=0 and F=-1, then the above equation becomes

(1)x^2 + 0xy + (1)y^2 + (0)x + (0)y +(-1) = 0

this simplifies to:

x^2 + y^2 = 1 which is the equation of a circle!

Or if you let A = 1, E=-1 and everything else zero, you get:

(1)x^2 + 0xy + (1)y^2 + (0)x+(-1)y + (0) = 0

this simplifies to y=x^2 which is a parabola!

So its all just a question of what coefficients should be zero!

**mikau**- Replies: 7

I started avoiding these forums because one member was making things a bit unpleasant for me but it appears that member isn't posting so much anymore.

Anyways! I got my associates degree in computer science and i should have one in math but I was one humanties class shy! but I changed schools and I'm continuing my Computeer Science education though i'm considering switching my major to math. I am unhappy with how little mathematics I've learned since I entered college, relative to before when I was self taught, and I'm trying to find time for more self study. These forums are also a great place to learn and practice math, by asking and answring questions, so I thought I should frequent them more!

anyways, good to be back everyone!

why is that, Simron? i don't use python..

**mikau**- Replies: 0

that i made up myself!

Whats a mathematicians favorite soda?

square root beer!

Whats a mathematicians favorite beer?

Natural Lagger!

Why did einstein cross the road?

well he insists the road crossed him!

Why did the mathematician cross the road?

he never crossed it, he approached it asymptotically

what is a helicopter bagel?

its like a plain bagel only it takes off vertically!

:]

I can easily say that at certain points of my life I spent over 20 hours a month playing games, and for the most part don't regret a minute of it.

No offense Ricky but I almost want to laugh! That's what, like 40 minutes a day? My parents only wish I played that little when I was younger!

If you consider that excessive than I'd say you have very good self control/discipline wish i had that

hey! no delegation!

nice! but you might as well move a big rock or that center of mass is gonna move awfully slow!