BEC'S methods questions

[bec]

Sorry, I didn't give you the rest of the question!

1. Let (a,b) be the point that is on both y = g(x) and y = g^-1(x). Find the coordinates of this point and prove that there are no other possible points.
so the thing here is, i have no problem proving that the point is at the origin (i solved g(x)=x, got 0 etc etc)...but i don't know how to prove that there are no other possible points.

...where g(x) = -(x+1)³+1

[/0]

Notice that when you graph a function and its inverse, all intersections must lie on the line y = x

Solve:









One solution is x = 0. To show that there are no other solutions, show that the discriminant of is less than 0. When this occurs there are no solutions to the quadratic. (The discriminant is )

[bec]

Thanks Divide

[bec]

How do you convert y = 2.4 x 2^x into the form y=e^{ct + d}?

[Mao]

hehe, nice attempt to restore your popularity [just joking!]

[bec]

thanks mao

hehe, nice attempt to restore your popularity [just joking!]

am i that transparent?

[bec]

I've come across a couple of Qs that I keep getting wrong...

A photocopier has a fault so that 15% of copies made are smudged. A person makes 80 copies and randomly selects 10 of the copies to distribute at a meeting. Let X=the number of smudged copies. Find Pr(X=3).
This is the way I was trying to work it out...what am I doing wrong? (btw when i've written 12C3 etc, i meant "combination")


Rashid and Khalil are friends who play a weekly game of sqash against each other. If Rashid wins the weekly game, there is a 40% chance that he will lose to Khalil the next week. If Khalil wins the weeky game, then there is a 45% chance that he will lose the next game.
If Khalil wins the game in the first week, what is the probability that he wins the next week, then Rashid wins the two weeks after that?
I get 0.2846, but the book says 0.1485....

Use the probability function below to find the mean, median and mode:
x      -2   0      2      5
p(x)  a  0.1  0.4  0.2

I thought the median would be 1, the mean of 0 and 2? (The answer is 2)


Ohhh and also, is 1 considered a prime number? (Edit: wikipedia told me. It's not.)
 
Thanks!

[Mao]

I've come across a couple of Qs that I keep getting wrong...

A photocopier has a fault so that 15% of copies made are smudged. A person makes 80 copies and randomly selects 10 of the copies to distribute at a meeting. Let X=the number of smudged copies. Find Pr(X=3).
This is the way I was trying to work it out...what am I doing wrong? (btw when i've written 12C3 etc, i meant "combination")

the only term wrong there is the 77C7. there are only 68 copies that are not smudged

Rashid and Khalil are friends who play a weekly game of sqash against each other. If Rashid wins the weekly game, there is a 40% chance that he will lose to Khalil the next week. If Khalil wins the weeky game, then there is a 45% chance that he will lose the next game.
If Khalil wins the game in the first week, what is the probability that he wins the next week, then Rashid wins the two weeks after that?
I get 0.2846, but the book says 0.1485....

assuming you've done a karnaugh table:

		Game 1 winner
		R	K
Game 2 winner	R	0.6	0.45
	K	0.4	0.55

probability that K wins the next week if K wins: 0.55
probability that R wins 2 weeks after if K wins (using tree diagram): 0.45*0.6 + 0.55*0.45 = 0.5175

final probability: 0.55*0.5175=0.2846
[it appears you are right, or we are both interpreting the question wrongly]

Use the probability function below to find the mean, median and mode:
x      -2   0      2      5
p(x)  a  0.1  0.4  0.2

I thought the median would be 1, the mean of 0 and 2? (The answer is 2)

if you had a dataset of {-2,0,2,5}, each with equal probability, then the median would be 1

more strictly, however, the median is the number separating the two halves of the probability distribution, i.e. where the probability gets to 0.5

This is occurs at the discrete value x=2

[bec]

Thanks mao - I never knew about the median thing, I'm glad I asked!

[bec]

Can anyone steady state matricies/probabilities to me?
I know how to form a transition matrix, and I understand that when a state matrix approaches a limit, the limit is the "steady state"...but I don't understand this:

"The steady state probabilities are such that:
Pr(X_n + 1 = 0) = Pr (X_n = 0) and Pr(X_n + 1 = 1) = Pr (X_n = 1)"

...and I also don't understand how it works with Markov chains.

Sorry, I know this is a really vague question but if anyone could give me a kind of outline of how all this works, that would be great. Otherwise I'll try and find a specific question that uses it.

Thanks

[Mao]

steady state matrices are simply, when the next state is the same as the current state.

when a markov chain is expressed as a transition matrix:

, where S are your states. (the transition matrix need not to be restricted to 2x2, but for this case it will do)

steady-state is such that , i.e. when probability stop changing. this is your "overall" probability in the "long run"



this gives us simultaneous equations



so, using either [1] or [2] in conjunction with [3] to solve, you can now find the probability of events at steady-state.

[bec]

hmmm
I don't really understand but I'll get there...

edit: i got there.

[bec]

A continuous random variable X has the probability density function f(x) = 1/x², x>1 (and 0 elsewhere)
Find exactly Pr(X>2|X=3)

The answer is 1/4. I would've thought it'd be 1? Since, given that X is 3, it has to be more than 2?

[Mao]

X EQUALS 3 ?!

its a continuous random variable... meaning .....
(if i remember correctly...)

[shinny]

I assume its a typo and that it should read Pr(X>2|X<3) since that'ld make alot more sense, since continuous random variables equal zero at any point as Mao showed. And hey, it gives the right answer too.