GR0568 Question 42

Forum for the GRE subject test in mathematics.
bonaluram
Posts: 7
Joined: Fri Oct 30, 2009 1:51 pm

GR0568 Question 42

Postby bonaluram » Fri Nov 06, 2009 2:30 pm

Hello,
Could you please help me about the following question :

Suppose X is a discrete random variable on the set of positive integers such that for each positive integer n, the probability that X = n is 1/(2^n). If Y is a random variable with the same probability distribution and X and Y are independent, what is the probability that the value of at least one of the variables X and Y is greater than 3?

(A) 1/64
(B) 15/64
(C) 1/4
(D) 3/8
(E) 4/9

The answer is B. Thank you all.

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: GR0568 Question 42

Postby joey » Fri Nov 06, 2009 3:38 pm

This is pretty straightforward when you know a few rules of probability. Write
\mbox{Pr}\left(X > 3 \mbox{ or } Y > 3\right) = 1 - \mbox{Pr}\left(X \le 3 \mbox{ and } Y \le 3\right)
Since X and Y are independent and identically distributed, this becomes
1 - \mbox{Pr}\left(X \le 3\right)\cdot\mbox{Pr}\left(Y \le 3\right)
= 1 - \mbox{Pr}\left(X \le 3\right)^2
= 1 - \left[\mbox{Pr}\left(X = 1\right) + \mbox{Pr}\left(X = 2\right) + \mbox{Pr}\left(X = 3\right)\right]^2

Then we just substitute and simplify.

bonaluram
Posts: 7
Joined: Fri Oct 30, 2009 1:51 pm

Re: GR0568 Question 42

Postby bonaluram » Fri Nov 06, 2009 3:45 pm

joey wrote:This is pretty straightforward when you know a few rules of probability. Write
\mbox{Pr}\left(X > 3 \mbox{ or } Y > 3\right) = 1 - \mbox{Pr}\left(X \le 3 \mbox{ and } Y \le 3\right)
Since X and Y are independent and identically distributed, this becomes
1 - \mbox{Pr}\left(X \le 3\right)\cdot\mbox{Pr}\left(Y \le 3\right)
= 1 - \mbox{Pr}\left(X \le 3\right)^2
= 1 - \left[\mbox{Pr}\left(X = 1\right) + \mbox{Pr}\left(X = 2\right) + \mbox{Pr}\left(X = 3\right)\right]^2

Then we just substitute and simplify.


Thank you so much joey, that really helped! =)

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: GR0568 Question 42

Postby joey » Fri Nov 06, 2009 4:07 pm

I'm glad to help! It beats the hell out of working, which is what I probably _should_ be doing right now...

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diogenes
Posts: 73
Joined: Sun Aug 31, 2008 9:31 pm

Re: GR0568 Question 42

Postby diogenes » Fri Nov 06, 2009 4:16 pm

joey wrote:I'm glad to help! It beats the hell out of working, which is what I probably _should_ be doing right now...



Heh...off today, but I understand. :lol:

bonaluram
Posts: 7
Joined: Fri Oct 30, 2009 1:51 pm

Re: GR0568 Question 42

Postby bonaluram » Fri Nov 06, 2009 4:38 pm

joey wrote:I'm glad to help! It beats the hell out of working, which is what I probably _should_ be doing right now...


I am so bored of this stuff, now i have less than 9 hours to the test. Last temptations :)

thmsrhn
Posts: 17
Joined: Fri Mar 26, 2010 7:18 am

Re: GR0568 Question 42

Postby thmsrhn » Fri Mar 26, 2010 4:25 pm

hey joey finish the whole problem please!!

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: GR0568 Question 42

Postby joey » Fri Mar 26, 2010 5:24 pm

We're told that \mbox{Pr}(X=n)=\frac{1}{2^n}. There's a joke about a mathematician, a physicist, and an engineer who are asked to put out a fire, but I won't get into it.




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