Page 1 of 1

GR0568 Question 42

Posted: Fri Nov 06, 2009 2:30 pm
by bonaluram
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.

Re: GR0568 Question 42

Posted: Fri Nov 06, 2009 3:38 pm
by joey
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.

Re: GR0568 Question 42

Posted: Fri Nov 06, 2009 3:45 pm
by bonaluram
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! =)

Re: GR0568 Question 42

Posted: Fri Nov 06, 2009 4:07 pm
by joey
I'm glad to help! It beats the hell out of working, which is what I probably _should_ be doing right now...

Re: GR0568 Question 42

Posted: Fri Nov 06, 2009 4:16 pm
by diogenes
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:

Re: GR0568 Question 42

Posted: Fri Nov 06, 2009 4:38 pm
by bonaluram
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 :)

Re: GR0568 Question 42

Posted: Fri Mar 26, 2010 4:25 pm
by thmsrhn
hey joey finish the whole problem please!!

Re: GR0568 Question 42

Posted: Fri Mar 26, 2010 5:24 pm
by joey
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.