<p>relevant to the dice question…</p>
<p>you have two coins. what’s the possibility of getting a head and a tail?
this is basically what everyone’s arguing over. does head+tail and tail+head count as one or two possibilities?</p>
<p>what was the answer for the line with a 30 degree incline? i used a 30, 60, 90 triangle and am not sure what the answer is</p>
<p>I agree with Owlehn. I did 1/6<em>1/6</em>2/6.</p>
<p>For the coin thing, it would be 2/2*1/2 = 1/2.</p>
<p>(HT, TH)/(HH, HT, TH, TT)</p>
<p>20 was an answer for 0<j<k<6. it was 4 x 5.</p>
<p>So far the only questions i think I missed was the Dice questions and that stupid inequality question. Hopefully this will be a 750 or better.</p>
<p>@spaceneil8 - see, but if you regard H+T and T+H as two separate possibilities, that means the number of possibilities for the dice sum to be 17 would be 3. which supports the 1/54 answer.</p>
<p>Does anyone remember the exact equation in the question with an answer like </p>
<p>x^2 < 3/5 y^2
or something like that? That’s the only question I’m concerned about, I really hope I didn’t misread something.</p>
<p>see, snoopi, heads/tails and tails/heads are the same so it is 1/108</p>
<p>^ so, you’re saying the possibility of getting H+T would be 1/3?</p>
<p>Nevermind ^^;</p>
<p>it would be x/sqrt(3) because tan(x)=y/x or rise/run, slope, so tan(30)=x/sqrt(3)</p>
<p>what would 10 omitted 4 wrong be ?</p>
<p>just want over a 610.</p>
<p>diablo: 620 ish? you’ll probably reach your goal</p>
<p>@firelight</p>
<p>thanks… i was really worried.</p>
<p>snoopi, well its hard to look at that way. this is the way i did the problem:</p>
<p>chance of getting 18; 1/6 <em>1/6</em> 1/6=1/216, no argument there, pretty simple.
chance of getting 17: now i put 1/6 <em>1/6</em> 1/6=1/216. my thinking here is that you need THREE specific numbers, two sixes and a five. the chance of getting a five is not any less than the chance of getting another six, the rolls are INDEPENDENT of eachother. the fact that you got two sixes means that you need either another five or six, so to simplify,
1/6 <em>1/6</em> 2/6=1/108.
if there is any flaw here, please clearly correct me.</p>
<p>^that’s EXACTLY the way i did. I also could not have said that better.</p>
<p>the first part is right but adding up to 17 there are 3 options, 665 656 and 566. same was as if you had two die and wanted to add up to 11 there would be two choices out of 36, 56 and 65 they’re different. so itd be 4 ways 666 665 656 566 out of 216=1/54</p>
<p>i dont understand how 566 665 656 are different? you need two sixes and a five, which doesnt affect the probability. the probability of getting two sixes and a five is the same as getting three sixes. you need three specific numbers</p>
<p>try this problem with two die and getting an 11, it’s essentially the same thing. if u write down all 36 possible ways to roll you’re gonna end up writing down both 56 and 65. and then you apply this to the 3 die problem</p>
<p>They are different. Think of it this way: if you do 6^3, that calculation indicates that there are 216 different possible combinations of numbers on the cubes. Given this and the fact that you can only have a sum of 17 in 3 ways (566, 665, 656), the probability of rolling a 17 is 3/216. If you apply the same concept to the chances of rolling a 18, since there is only one possible combination that will yield a sum of 18, you get 1/216. Adding these probabilities together and simplifying, you get 1/54.</p>