Another set (4) of very common questions

 Posts: 18
 Joined: Sun Oct 07, 2007 1:49 pm
Another set (4) of very common questions
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(xh)]/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...
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(xh)]/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...
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_tI)/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(xu)}/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(xu)}/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(xu)}/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 nonlinear 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!
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_tI)/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(xu)}/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(xu)}/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(xu)}/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 nonlinear 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!
In my opinion, something is definitely wrong with this question!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)
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?
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
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
Yeah, I totally agree here that in general P(A)=P(B) does not imply such conclusion. But, you forgot thatP(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)..
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.
It is easy to see that A=B is not necesary for the condition to hold,I can be wrong but since
P(A) = P(B) = P(A intersects B) < 1
it must be
A = B !
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.
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
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

 Posts: 8
 Joined: Tue Mar 04, 2008 1:02 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.
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.