<p>ah. well, i don't remember ever learning that. evil problem.</p>
<p>Actually, now that I think about it, Weren't the digits (1-9)? That would leave only nine choices, but I don't think 1* 26<em>26</em>9<em>9</em>9 wasn't a choice. Maybe it was??</p>
<p>it said "each of the ten digits", which means 0-9</p>
<p>no, there are ten digits. do you want me to list them out for you?</p>
<p>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</p>
<p>the first digit had to be the number "8" so therefore there is only ONE choice. and since numbers and letters can repeat, it would be:</p>
<p>(1)(26)(26)(10)(10)(10)</p>
<p>capiche?</p>
<p>Thanks guys
(=</p>
<p>probability with dice??</p>
<p>what were the answers to the first 2 english questions!?! they were the only ones i had trouble with. The first was was deciding whether or not to combine the first two sentances and i forget what the second one was.?!?!?</p>
<p>5/36 for the dice thing. At least, that is what I got. OMG Sophie, the first passage gave me so much trouble. After that, it was ok. I would comment about my answers, but I don't remember them.</p>
<p>5/36 is what i put as well</p>
<p>yup inguyen... apart from the first 2 questions on the english... the test was quite easy :)</p>
<p>Yep, I got 5/36 also.</p>
<p>The dice problem was 1/6. The combinations of upward-facing sides that add to 8 are:</p>
<p>6,2
5,3
4,4</p>
<p>The total number of combinations is 6^2/2 = 18 (36 would count each twice). Thus, the probability is 3/18 = 1/6.</p>
<p>Well I'm REALLY smart. I think I may have counted 7 and 1. Oiiii. hahaha</p>
<p>Wouldn't you count each twice though, because dice 1 could be a 6, dice 2 could be a 2... or the other way around, but there's two chances that that combo could happen, no one.</p>
<p>you forgot to switch the dice for (6,2) and (5,3). resulting in 5 combinations out of a total 36</p>
<p>dice 1 - 6
dice 2 - 2</p>
<p>dice 1 - 2
dice 2 - 6</p>
<p>dice 1 - 5
dice 2 - 3</p>
<p>dice 1 - 3
dice 2 - 5</p>
<p>dice 1 - 4
dice 2 - 4</p>
<p>Counting each possible combination twice would require you to count each combination adding to 8 twice also. If we said there were 36 combinations, then our list of combinations adding to 8 would necessarily become:</p>
<p>6,2
5,3
4,4
4,4 (Imagine the two 4s swapped places)
3,5
2,6</p>
<p>That's 6/36, or 1/6.</p>
<p>yeah, don't count the 4's switching places. 5/36</p>
<p>oh oh, i think i counted (7,1) and (1,7), goddammit, say goodbye to 36 math :'(</p>
<p><em>sigh</em> I think you're right Tickitata. :(</p>
<p>All the possibilities are:</p>
<p>(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) <a href="2,%206">b</a><a href="3,%201">/b</a> (3, 2) (3, 3) (3, 4) <a href="3,%205">b</a><a href="3,%206">/b</a>
(4, 1) (4, 2) (4, 3) <a href="4,%204">b</a><a href="4,%205">/b</a> (4, 6)
(5, 1) (5, 2) <a href="5,%203">b</a><a href="5,%204">/b</a> (5, 5) (5, 6)
(6, 1) <a href="6,%202">b</a><a href="6,%203">/b</a> (6, 4) (6, 5) (6, 6)</p>
<p>The ones adding to 8 are bolded. There are five out of a total of 36.</p>
<p>So now I know of two math questions that I missed definitely...</p>
<p>heh, combinations are a b***.</p>
<p>funny story. last year, during a math test, it was a simple combination problem for a license plate (first few are numbers, last few are letters. something like that). and i was like "..wait there's 26 letters in the alphabet, right?" and to double check and make sure i don't mess up on something that stupid, i sang the song in my head, counting each letter</p>
<p>well, at the end i counted 27. (x, y, AND, z) i was like "wow i'm glad i double checked"</p>
<p>oops.</p>
<p>yeah for that dice i also got 1/6</p>
<p>basically you know on the first you can get 6/6 but on the second you have to hit 1/6 so </p>
<p>6<em>1/6</em>6 = 6/36 = 1/6</p>