<p>In a stack of six cards, each card is labeled with a different integer 0 through 5. If two cards are selected at random without replacement, what is the probability that their sum will be 3?</p>
<p>well... i got the answer (2/15) by just listing out the possibilities... but i was wondering if there was a way to do it by using Combinations or something.</p>
<h1>of different of possibilities of picking 2 cards.</h1>
<p>6<em>P</em>2 = 30</p>
<p>Possibilities that make the sum 3.</p>
<p>0 3
3 0
1 2
2 1</p>
<p>Therefore, possibility of picking 2 cards that add up to 3 is 4/30, which is equivalent to 2/15.</p>
<p>6 cards</p>
<p>0 1 2 3 4 5</p>
<p>total probability
5+4+3+2+1= 15
it's PAIRS of SUMS (the 6 *5 thing doesn't work)</p>
<p>total probablility of picking at random two cards
3 and 0
1 and 2
it's PAIRS of SUMS so we cannot switch orders by using 0 and 3, 2 and 1. 3 and 0 as opposed to 0 and 3 are essentially the same pair and goes for the same with the 1 and 2 vs. 2 and 1.</p>
<p>2/15 is the answer</p>
<p>umm....i don't know what you did with that problem..im surprised you got the right answer lol</p>
<p>what do you mean by pairs of sums. -_-; that doesn't really matter. because there are 4 ways to obtain 2 cards that sum up to 3.</p>
<p>1) pick 0 , then pick 3
2) pick 3, then pick 0
3) pick 1, then pick 2
4) pick 2, then pick 1</p>
<p>and there are 6*5 different possibilties of picking two cards. </p>
<p>you think that picking 0 and then picking 3 is the same as picking 3 and then picking 0. well..they are NOT.</p>
<p>If we do 6 *5
it is like this picking 5 and 4, yet we repeat again by picking 4 and 5. It wants us to find the pair and 4 and 5/ 5 and 4 are the same pair.</p>
<p>5+4+3+2+1
which means:
picking
5 4
5 3
5 2
5 1
5 0
4 3
4 2
4 1
4 0
3 2
3 1
3 0
2 1
2 0</p>
<p>which equals to 15 DIFFERENT pairs.</p>
<p>picking
5 4</p>
<p>is the same PAIR as picking
4 5</p>
<p>wow, didn't see anyone get so worked up on a stupid math problem. posting 3 times in a row, repeating the same thing over and over again lol. </p>
<p>we got the same answer anyway. YOU didn't care about the order, while I did.</p>
<p>there is never only one way to solve a problem. i probably wouldn't have posted if your post didn't have the "im right, you're wrong" attitude.</p>
<p>and you said in your first post,
it's pairs of sums so we "CANNOT" switch orders by using 0 and 3, 2 and 1. 3 and 0 as opposed to 0 and 3 are essentially the same pair and goes for the same with the 1 and 2 vs. 2 and 1.</p>
<p>well i guess we can, cuz i got the same answer as you, and my method is not wrong. maybe a little longer than yours, but not WRONG.</p>
<p>I'm sorry that I offended you, and I did do it the same way as you did it at first but I kept anaylzing it and found this way. The problem states "if two cards are selected at random..." so we would have to pick 2 at the same time. That was all I wanted to prove that 2 cards are being picked at the same time.</p>
<p>heh, I apologize also. After being yelled at for about 30 min, my mood wasn't exactly great =P. Ya, picking 2 cards randomly surely makes your method work better than mine. I just cared about the order because it said without replacement, which means that they pick one, and then pick another lol. but your method is more efficient than mine tho.</p>
<p>Thanks coolness_rookie, btw is this question from the CB?</p>
<p>I have no idea. (don't even know what CB is, lol.) in fact, I haven't even started preparing for SAT's or anything at all. haha. i know im in some trouble, since i'm gonna write the PSAT in october and SAT in like december :P:P</p>
<p>I'm also going for the PSAT this oct., CB stands for College Board.</p>
<p>what grade are you in, retribution?</p>
<p>I'm going to be a junior, what about you?</p>
<p>hahah same here =)</p>
<p>if you don't mind me asking,</p>
<p>1) where do you live in?
2) what colleges are in thinking of applying to? </p>
<p>=) this is kind of interesting. its like....talking to a friend on a really ancient version of MSN. haha</p>
<p>Do you want to talk on aim or msn?</p>
<p>I solved this problem without making order matter. All you have to do is (6 C 2), which equals 6!/(4!<em>2!) or 6</em>5/2 = 15. Then since we didn't use order, the only 2 ways of getting a pair adding to 3 is (0,3) and (1,2). So the probability is 2/15.</p>