9367 #40

Forum for the GRE subject test in mathematics.
origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

9367 #40

Postby origin415 » Thu Nov 05, 2009 1:17 pm

If x, y, and z are selected independently and at random from the interval [0, 1], then the probability that x >= yz is
A) 3/4
B) 2/3
C) 1/2
D) 1/3
E) 1/4

The answer is A
I got it right, but couldn't figure out how to actually calculate it, I just guessed that it must be pretty high. Its obviously greater than 1/2, but my limited knowledge of probability theory prevented me from getting any further.

Thanks.

mtey
Posts: 22
Joined: Tue Oct 27, 2009 12:30 pm

Re: 9367 #40

Postby mtey » Thu Nov 05, 2009 1:28 pm

the triple density function

f_{XYZ}(x,y,z) = 1 since they are independent. So the probability is

\displaystyle \int \int \int_{D}1dxdydz where
D = \begin{cases}<br />yz \leq x \leq 1 \\ 0 \leq y \leq 1 \\ 0 \leq z \leq 1 \\ \end{cases}

origin415
Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: 9367 #40

Postby origin415 » Fri Nov 06, 2009 12:16 am

Ah, I was trying to do an integral, 0 < x, y < 1, 0 < z < x/y, which had quite a bit of complications with y going to zero and all. The other direction makes a lot more sense, thanks.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 #40

Postby yoyobarn » Fri Dec 16, 2011 9:19 am

mtey wrote:the triple density function

f_{XYZ}(x,y,z) = 1 since they are independent. So the probability is

\displaystyle \int \int \int_{D}1dxdydz where
D = \begin{cases}<br />yz \leq x \leq 1 \\ 0 \leq y \leq 1 \\ 0 \leq z \leq 1 \\ \end{cases}


Can someone elaborate more on this?
Is there a way without doing triple integration?

Thanks a lot.

goombayao
Posts: 53
Joined: Sun Oct 16, 2011 9:17 am

Re: 9367 #40

Postby goombayao » Fri Dec 16, 2011 8:08 pm

E(yz) = E(y)*E(z) = 1/2 * 1/2 = 1/4

P(x > 1/4) = 3/4

yoyostein
Posts: 36
Joined: Tue Feb 28, 2012 12:14 am

Re: 9367 #40

Postby yoyostein » Tue Apr 10, 2012 12:16 am

E(yz) = E(y)*E(z) = 1/2 * 1/2 = 1/4

P(x > 1/4) = 3/4


Can anyone explain why the above method works?

If I consider the expectation of X instead,
E(X)=1/2

P(yz > 1/2) != 1/2 ?


Sincere thanks for the clarification.

Vince
Posts: 4
Joined: Sat Mar 17, 2012 4:36 pm

Re: 9367 #40

Postby Vince » Tue Apr 10, 2012 6:28 am

yoyostein wrote:
E(yz) = E(y)*E(z) = 1/2 * 1/2 = 1/4

P(x > 1/4) = 3/4


Can anyone explain why the above method works?

If I consider the expectation of X instead,
E(X)=1/2

P(yz > 1/2) != 1/2 ?


Sincere thanks for the clarification.



Since y and z are independent, so the mean of a product is a product of their means. Therefore, E(zy) = E(z)*E(y)=1/2*1/2=1/4.
In order for x > yz => x should be > 1/4. Again, x is chosen randomly(!), so we can figure out this probability easily: (1-1/4)/(1-0)=3/4.

Your method is incorrect, because using this solution you give an answer to another question.




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