answer:Instead of directly finding the probability that at least two faces match, we can find the probability that no faces match and then subtract the result from 1. The results on the three dice are independent of each other, so we compute the probability for each die and then multiply the probabilities. The first die does not have to be a particular number. There are 6 possible numbers, but any number will work, so the probability is frac{6}{6}=1. In order for the second die to have a different number from the first, there are 5 other numbers out of the 6 possible outcomes, so the probability is frac{5}{6}. For the third die to have a different number from the first and second, there are 4 other numbers out of 6 possible outcomes, so the probability is frac{4}{6}=frac{2}{3}. The probability that no faces match is 1timesfrac{5}{6}timesfrac{2}{3}=frac{5}{9}. That means the probability that at least two faces match is 1-frac{5}{9}=frac{4}{9}.

answer:I notice that the word CEASE has two E's that are indistinguishable, so I can't just use the formula for permutations of n distinct objects. Instead, I need to account for the fact that swapping the two E's does not create a new arrangement. One way to do this is to divide by the number of ways to arrange the two E's among themselves, which is 2!. So, the number of ways to arrange the letters of CEASE is n! / 2!, where n is the number of letters in the word. In this case, n = 5, so the answer is 5! / 2! = 120 / 2 = 60.

answer:If a triangle with sides of length 1, x, and y exists, the triangle inequality must be satisfied, which states that x+y>1, 1+x>y, and 1+y>x. We can draw a plane with x and y axes and shade in the area where all of these inequalities are satisfied. [asy] draw((0,0)--(3,0)--(3,3)--(0,3)); draw((0,0)--(0,3)); label("x",(3,0),S); label("y",(0,3),W); fill((1,0)--(3,2)--(3,3)--(2,3)--(0,1)--cycle,gray(.7)); draw((1,-.1)--(1,.1)); draw((2,-.1)--(2,.1)); draw((.1,1)--(-.1,1)); draw((.1,2)--(-.1,2)); draw((1,0)--(0,1)); draw((1,0)--(3,2)); draw((0,1)--(2,3)); [/asy] The total area of the square is 3^2=9. The area of the unshaded region is 2^2+1/2=9/2. Thus, the shaded area is 9/2 and the probability that such a triangle exists is (9/2)/9=frac{1}{2}.

answer:This problem involves counting circular permutations, which are different from linear permutations because there is no fixed starting point or direction on a circle. One way to approach this problem is to fix one of the beads as a reference point, and then arrange the remaining four beads around it in a clockwise or counterclockwise order. This would give me 4! = 24 ways to arrange the four beads, but this would overcount by a factor of 2, because flipping the bracelet over would give me the same arrangement in the opposite order. So I need to divide 4! by 2 to get the correct number of circular permutations. Alternatively, I could use a formula for circular permutations, which is (n-1)!/2, where n is the number of objects to arrange. Either way, I get the same answer: 4! / 2 = (5-1)! / 2 = 12.