<p>there are 3 6-sided dice. so lets take one dice. the chances of getting any number on a dice is 1/6. we have to find the chances of getting 18…and with a 6-sided die thats only possible if we roll all 6’s. the chances of getting a 6 with one dice is 1/6, so the chances of getting 3 6’s is 1/6 * 1/6 * 1/6, or 1/216. then we need to find the chances of getting a 17…with three dice there are 3 different combinations that add to 17. a 5,6,6; a 6,5,6; and a 6,6,5. each of these has a 1/216 chance of happening (like my explanation above for each number), so since this can happen three times it wud b 1/216 +1/216 + 1/216 or 3/216. add this to the 18 we also have to get and u get 3/216 + 1/216 which wud b 4/216…or…</p>
<p>@1x612nt13 I agree with all of those (except for 1/54. I put 1/108, but I’ve accepted my fate.) And no way -1 equals 760. There hasn’t been a curve that harsh ever, plus I still don’t even think this was that easy of an exam. Back in November last year, it went 800-770-750 and I thought that today’s was harder. I’m thinking something along the lines of 800-780-760.</p>
<p>Thank you all for the clarification. I just constructed it with a ruler & compass, and a 6-8-9 triangle would indeed be acute. I should have been paying attention to the pythagorean theorem for the side lengths.</p>
<p>By the way - what were some problems from the experimental math section?</p>