Another set (4) of very common questions

Forum for the GRE subject test in mathematics.
fullofquestions
Posts: 18
Joined: Sun Oct 07, 2007 1:49 pm

Another set (4) of very common questions

Postby fullofquestions » Wed Oct 31, 2007 11:08 pm

This is another set of questions that I've seen often and think are worthwhile covering. Any help is appreciated. Thank you.

I
If A and B are events in a probability space such that 0 < P(A) = P(B) = P(A intersect B) < 1, which of the following cannot be true?

ans a) A and B are independent
b) A is a proper subset of B
c) A != B
d) A intersect B = A union B
e) P(A)P(B) < P(A intersect B)

I thought the answer would be C since P(A) = P(B) = P(A intersect B)...


II This type of question is very very common
If x, y and z are selected independently and at random from the interval [0,1], then probability that x >= yz is

ans a) 3/4
b) 2/3
c) 1/2
d) 1/3
e) 1/4

Intuitively I know that yz is most likely smaller than x but I cannot go beyond saying that the answer is over 1/2.


III Another very common type
If x is a real number and P is a polynomial, then lim h->0 [P(x + 3h) + P(x - 3h) - 2P(x)]/h^2 =

a) 0
b) 6P'(x)
c) 3P''(x)
ans d) 9P''(x)
e) infinity

I saw another problem where they asked the same except the limit was lim h->0 [P(x+h) - P(x-h)]/h and I intuitively answered correctly, that the answer was 2P'(x). What is the procedure for this type of question?


IV
Consider the sytem of equations

ax^2 + by^3 = c
dx^2 + ey^3 = f

Where a, b, c, d, e and f are real constants and ae != bd. The maximum possible number of real solutions (x,y) of the system is

a) none
b) one
ans c) two
d) three
e) five

I thought a system of equations would have only 0, 1 or infinitely many solutions...

mamal
Posts: 11
Joined: Tue Oct 30, 2007 3:02 am

Postby mamal » Thu Nov 01, 2007 4:36 am

I)
A and B are independent iff: P(A intersect B)=P(A)P(B). However in this question. P(A intersect B)=P(A) != P(A)P(B)=P(A)^2. since P(A) != 0 or 1.

II)
since x,y,z are all less than unity and greater than zero, the total volume where x,y,z may be selected from is V_t=1. now we should calculate the volume between these lines and serface x=yz. for that should do the integral as I= INTEG[INTEG[yz dydz](0 to 1)](0 to 1)=INTEG[y dy](0 to 1)*INTEG[z dz](0 to 1)=1/4. now P=(V_t-I)/V_t=3/4

III)
The procedure is easy. You need to recognize the definition of a derivative. Let us rewrite the problem in the following form.
L=Lim u->0 9*{f(x+u)-2*f(x)+f(x-u)}/u^2, where u=3*h. Now noting that f'(x)=lim u->0 {f(x+u)-f(x)}/u=lim u->0 {f(x)-f(x-u)}/u, we can write that f''(x)=[f'(x)]'=lim u->0 {f'(x+u)-f'(x)}/u=lim u->0 {f(x+u)-2f(x)-f(x-u)}/u^2. Thus for this problem L=9*P''(x)

IV)
You are partially right. I mean a system of LINEAR equations have 0, 1 or infinite soultions, however this set of equations are non-linear in x and y. but what if we substitude u=x^2 and v=y^3, this way the equations become linear in u and v and since ae !=bd, we have one set of (u,v) that satisfies the equations. however since u=x^2 and v=y^3 we have TWO set of solutions in terms of x,y and they are (SQRT(u), v^(1/3)) and (-SQRT(u), y^(1/3))

Hope this has helped you!

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lime
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Joined: Tue Dec 04, 2007 2:11 am

Postby lime » Tue Feb 19, 2008 7:00 pm

I
If A and B are events in a probability space such that 0 < P(A) = P(B) = P(A intersect B) < 1, which of the following cannot be true?
a) A and B are independent
b) A is a proper subset of B
c) A != B
d) A intersect B = A union B
e) P(A)P(B) < P(A intersect B)

In my opinion, something is definitely wrong with this question! :!:
I can be wrong but since
P(A) = P(B) = P(A intersects B) < 1
it must be
A = B !
Which also implies that variants b), c) and e) also must be wrong.
Colleague, what do you think?

Pablo
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Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Thu Feb 28, 2008 12:24 am

P(A)=P(B) does not imply A=B, for example A= you get 1 in a dice, B= you get 2, P(A)=P(B)..

Since P(A and B)=P(A)*P(B) IF A and B are independent, the equalities does not hold if they are strictly positive.


For problem 2:

The only way that yz<x is that y<x or z<x, you just need one to be less than x, so you get P(x>yx)=P(x>y)+(x>z)-P(x>y)*P(x>z)

if you draw the unit square in the plane, you'll see that P(x>y)=1/2
and doing the arithmetic we get 3/4

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lime
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Postby lime » Thu Feb 28, 2008 1:22 am

P(A)=P(B) does not imply A=B, for example A= you get 1 in a dice, B= you get 2, P(A)=P(B)..

Yeah, I totally agree here that in general P(A)=P(B) does not imply such conclusion. But, you forgot that
P(A)=P(B)=P(A intersects B)>0
For your example with a dice P(A intersects B) would be equal to 0, which contradicts to the conditions.

Pablo
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Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Thu Feb 28, 2008 11:07 am

I can be wrong but since
P(A) = P(B) = P(A intersects B) < 1
it must be
A = B !


It is easy to see that A=B is not necesary for the condition to hold,

In General, let P(A)>0, we define B=A union C, with C an nonempty set with P(C)=0

We have P(A)=P(B)=P(A intersection B), But A!=B,

Then we see that b) holds, which implies c), from the condition of the problem e) always hold. Now, if we let A=B then d) is true. The one that is necesarily false is the independence, in that case:

0<P(A)=P(B)=P(A intersection B)=P(A)P(B)<1 # Absurd.

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lime
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Postby lime » Thu Feb 28, 2008 11:57 am

C a nonempty set with P(C)=0

Sounds great! And I see that this is actually the point of my incomprehension. Can you give an example of such set?

Pablo
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Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Thu Feb 28, 2008 10:04 pm

You can have nonempty sets with measure zero, (like rationals under lebesgue)

Let X ~ Uniformly over the interval [0,1]

let A=[0,1/2], and C={ x elements of A | x is rational}

C is not empty, indeed it has infinitely (but countable) many elements, but P(C)=0,

in this case we say that Xj is the set that contains only the j'th rational (we can order them as in cantor diagonals.. )

P(countable sum of disjoint Xj)=countable sum P(Xj)
But the probability of a point is 0, so P(C)= sum of zeros=0

Pablo
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Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Thu Feb 28, 2008 10:09 pm

for a more mundane example, supose that we have a biased dice such that we can never get 1 or 2 in a throw, and supose that the other numbers are equaly likely to apear:

the set A={3}, let C={1,2}, and B= A union C={1,2,3}

P(C)=0
0<P(A)=P(B)=P(A intersection C)=1/4<1

Pablo
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Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Thu Feb 28, 2008 10:14 pm

By the way lime, can you recomend me introductory books on abstract algebra and topology? Something for a crash course.

Pablo
Posts: 6
Joined: Thu Feb 28, 2008 12:05 am

Postby Pablo » Mon Mar 03, 2008 2:36 pm

For problem 3:

[P(x+3h)+P(X-3h)-2P(x)]/h^2, numerator and denominator ->0 as h->

Apply l´Hôpital 2 times:

First time [3P'(x+3h)-3P'(x-3h)]/2h

Second time [9P''(X+3h)+9P''(x-3h)]/2 and now take the limit to get

9P''(X)

ralphhumacho
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Joined: Tue Mar 04, 2008 1:02 am

Postby ralphhumacho » Tue Mar 04, 2008 1:23 am

Not too beat a dead horse, after seeing some unorthodox methods of solving this problem, here is a quick way to do it.

We know that:
E(x)=E(y)=E(z)=.5

E(yz)=E(y)E(z)=(.5)(.5)=.25
(This is true by some Theorem I forgot...)

Pr(x>.25) = .75, since x is between 0,1.

I took a course in prob/stats a while back, and i believe this is how we were taught to approach the problem.




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