## 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

Hello,

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

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

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

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

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

### Re: GR0568 Question 42

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.

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

### Re: GR0568 Question 42

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

hey joey finish the whole problem please!!

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

### Re: GR0568 Question 42

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.