What is the probability that the sum is 7 A 1 9 B 1 8 C 1 6 D 2 11 E 1 5 2009

# What is the probability that the sum is 7 a 1 9 b 1 8

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the numbers on the two top faces are added. What is the probability that the sum is 7 ? (A)19(B)18(C)16(D)211(E)15 2009 AMC 10 A, Problem #22“Theprocessofrandomlypickingthenumbers,randomly axing them to the dice, rolling the dice, andadding the top numbers is equivalent to picking two ofthe twelve numbers at random and adding them.” Solution Answer (D): Suppose that the two dice originally had the numbers 1, 2, 3, 4, 5, 6 and 1 0 , 2 0 , 3 0 , 4 0 , 5 0 , 6 0 , respectively. The process of randomly picking the numbers, randomly affixing them to the dice, rolling the dice, and adding the top numbers is equivalent to picking two of the twelve numbers at random and adding them. There are ( 12 2 ) = 66 sets of two elements taken from S = { 1 , 1 0 , 2 , 2 0 , 3 , 3 0 , 4 , 4 0 , 5 , 5 0 , 6 , 6 0 } . There are 4 ways to use a 1 and 6 to obtain 7, namely, { 1 , 6 } , { 1 , 6 0 } , { 1 0 , 6 } , and { 1 0 , 6 0 } . Similarly there are 4 ways to obtain the sum of 7 using a 2 and 5, and 4 ways using a 3 and 4. Hence there are 12 pairs taken from S whose sum is 7. Therefore the requested probability is 12 66 = 2 11 . Difficulty: Hard NCTM Standard: Data Analysis and Probability Standard: understand and apply basic concepts of probability. Mathworld.com Classification: Probability and Statistics > Probability > Probability Convex quadrilateral ABCD has AB = 9 and CD = 12 . Diagonals AC and BD intersect at E , AC = 14 , and 4 AED and 4 BEC have equal areas. What is AE ? 2009 AMC 10 A, Problem #232009 AMC 12 A, Problem #20ABkCD.” Solution Answer (E): Because 4 AED and 4 BEC have equal areas, so do 4 ACD and 4 BCD . Side CD is common to 4 ACD and 4 BCD , so the altitudes from A and B to CD have the same length. Thus AB k CD , so 4 ABE is similar to 4 CDE with similarity ratio AE EC = AB CD = 9 12 = 3 4 . Let AE = 3 x and EC = 4 x . Then 7 x = AE + EC = AC = 14 , so x = 2 , and AE = 3 x = 6 . A B C D E 9 12 Difficulty: Hard NCTM Standard: Geometry Standard: explore relationships (including congruence and similarity) among classes of two- and three-dimensional geometric objects, make and test conjectures about them, and solve problems involving them. Mathworld.com Classification: Geometry > Plane Geometry > Quadrilaterals > Quadrilateral Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube? 2009 AMC 10 A, Problem #24“A plane that intersects at least three vertices of acube either cuts into the cube or is coplanar with acube face.” Solution Answer (C): A plane that intersects at least three vertices of a cube either cuts into the cube or is coplanar with a cube face. Therefore the three randomly chosen vertices result in a plane that does not contain points inside the cube if and only if the three vertices come from the same face of the cube. There are 6 cube faces, so the number of ways to choose three vertices on the same cube face is 6 · ( 4 3 ) = 24 . The total number of ways to choose the distinct vertices without restriction is ( 8 3 ) = 56 . Hence the probability is 1 - 24 56 = 4 7 . Difficulty: Hard NCTM Standard: Geometry Standard: analyze properties and determine attributes of two- and three-dimensional objects. Mathworld.com Classification: Geometry > Solid Geometry > Polyhedra > Cubes > Cube For k > 0 , let I k = 10 . . . 064 , where there are k zeros between the 1 and the 6. Let N ( k ) be the number of factors of 2 in the prime factorization of I k . What is #### You've reached the end of your free preview.

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