<p>how do i solve this sh it?</p>
<p>Set X= {1,3,5}
Set Y= {2,4,6}</p>
<p>If x and y are to be selected from random sets X and Y,respectively, what is the probability that 2x/y will be a member of set X?</p>
<p>it was the very last question in the math section of a practice test that i just took. It's the only one i missed. had 3 minutes to do it but didn't even know where to start.</p>
<p>Possible combinations of 2x/y: 2(1)/2, 2(1)/4, 2(1)/6, 2(3)/2, 2(3)/4, 2(3)/6, 2(5)/2, 2(5)/4, 2(5)/6</p>
<p>That's a total of 9 possible combinations, or 3^2.</p>
<p>And the only instances where 2x/y={1,3,5} are:
-2(1)/2=1
-2(3)/2=3
-2(5)/2=5</p>
<p>Which is a total of 3.</p>
<p>Thus, 3/9=1/3 probability. I believe that is how it's done. Hope this helps.</p>
<p>Ok now i feel like an a ss...I didn't understand the question at all. at first i was wondering how one's supposed to pick x and y from sets that dont have 'x' and 'y' in them. But little x represents any number picked from the X set and little y represents any number picked from the Y set. thanks--</p>
<p>Set X= {1,3,5}
Set Y= {2,4,6}</p>
<p>If x and y are to be selected from random sets X and Y,respectively, what is the probability that 2x/y will be a member of set X?</p>
<p>(2<em>1)/2 =1
(2</em>3)/2 =3
(2<em>5)/2 = 5
and
(2</em>3)/6 =1</p>
<p>four good choices out nine.</p>
<p>Wow, I totally overlooked that 2(3)/6 one. Thanks for the correction, my$0.02. And nice username. :)</p>