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9367 #40

Posted: Thu Nov 05, 2009 1:17 pm
by origin415
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.

Re: 9367 #40

Posted: Thu Nov 05, 2009 1:28 pm
by mtey
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
$$D = \begin{cases}
yz \leq x \leq 1 \\ 0 \leq y \leq 1 \\ 0 \leq z \leq 1 \\ \end{cases}$$

Re: 9367 #40

Posted: Fri Nov 06, 2009 12:16 am
by origin415
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.

Re: 9367 #40

Posted: Fri Dec 16, 2011 9:19 am
by yoyobarn
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
$$D = \begin{cases}
yz \leq x \leq 1 \\ 0 \leq y \leq 1 \\ 0 \leq z \leq 1 \\ \end{cases}$$
Can someone elaborate more on this?
Is there a way without doing triple integration?

Thanks a lot.

Re: 9367 #40

Posted: Fri Dec 16, 2011 8:08 pm
by goombayao
E(yz) = E(y)*E(z) = 1/2 * 1/2 = 1/4

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

Re: 9367 #40

Posted: Tue Apr 10, 2012 12:16 am
by yoyostein
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.

Re: 9367 #40

Posted: Tue Apr 10, 2012 6:28 am
by Vince
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.