GR0568 Question 42

[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.

[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.

[bonaluram]

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]

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

[diogenes]

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]

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]

hey joey finish the whole problem please!!

[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.