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• 2018 AIME II Problems/Problem 13
  ...she rolls <math>1-2-3</math> in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is <math>\dfrac{m}{n}</ma Let <math>P_n</math> be the probability of getting consecutive <math>1,2,3</math> rolls in <math>n</math> rolls and
  11 KB (1,860 words) - 13:12, 24 January 2024
• 2002 Indonesia MO Problems/Problem 2
  ...hen the product of the <math>5</math> numbers shown are calculated. Which probability is bigger; the product is <math>180</math> or the product is <math>144</mat ...ice, and <math>e</math> be the roll of the fifth dice. To calculate which probability is bigger, find the number of ways to roll dice that result in the two want
  3 KB (406 words) - 00:38, 31 January 2024
• 2008 iTest Problems/Problem 49
  ...students around the table are equally likely, let <math>m/n</math> be the probability that the yarns intersect, where <math>m</math> and <math>n</math> are relat [[Category:Intermediate Probability Problems]]
  1 KB (240 words) - 00:03, 6 August 2018
• 2008 iTest Problems/Problem 66
  ...math>AB</math> is <math>4</math> feet in length, let <math>p</math> be the probability that the number of feet in the length of <math>AD</math> is less than <math ...> Since <math>[ACD'] = 4-2\sqrt3</math> and <math>[ACB] = 2,</math> the [[probability]] <math>p</math> is equal to <math>2-\sqrt3.</math>
  2 KB (359 words) - 09:39, 8 August 2018
• 2008 iTest Problems/Problem 44
  ...> are relatively prime positive integers such that <math>a/b</math> is the probability that the product of the two terms is positive, find the value of <math>a+b< ...th> positive terms. We can use [[complementary counting]] to calculate the probability of getting a negative product because it is easier than adding the probabil
  2 KB (284 words) - 23:27, 6 May 2019
• 2016 UMO Problems/Problem 2
  Four fair six-sided dice are rolled. What is the probability that they can be divided into two pairs which sum to the same value? For ex <cmath>\textbf{Probability}=\frac{\textbf{Favorable}}{\textbf{Total}}.</cmath>
  6 KB (950 words) - 14:53, 7 December 2021
• 2008 iTest Problems/Problem 26
  The proof goes as follows: the probability of rolling n on roll 1 is <math>(\frac{1}{12})</math>. The probability of rolling n on roll 2 (but not roll 1) is <math>(\frac{11}{12})\cdot (\fra
  3 KB (461 words) - 13:09, 2 January 2020
• 2019 AIME I Problems/Problem 5
  ...ability <math>\frac{1}{3}</math>, independently of its previous moves. The probability that it will hit the coordinate axes at <math>(0,0)</math> is <math>\frac{m ...dot \frac{1}{729}=\frac{245}{729}</math> to get to <math>(1,1)</math>. The probability of reaching <math>(0,0)</math> is <math>\frac{245}{3^7}</math>. This gives
  4 KB (668 words) - 15:51, 25 February 2021
• 2019 AIME II Problems/Problem 4
  A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is <math>\t Case 1 (easy): Four 5's are rolled. This has probability <math>\frac{1}{6^4}</math> of occurring.
  19 KB (2,942 words) - 21:22, 21 January 2024
• 2006 Indonesia MO Problems/Problem 6
  ...4103200</math> phone numbers with at most five distinct digits. Thus, the probability that Mr. Edy gets a phone number with at most five distinct digits is <math ...math> phone numbers that satisfies the conditions, which confirms that the probability is <math>\boxed{0.41032}</math>.
  5 KB (819 words) - 16:21, 20 December 2019
• 2020 AIME I Problems/Problem 9
  ...ath>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> di ...eeds similarly for a result of <math>\dbinom{12}{3}</math>. Therefore, the probability of choosing three such factors is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom
  6 KB (1,036 words) - 00:59, 25 February 2021
• GMAAS
  ...herefore, I proved that you cannot use the Almighty Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Almighty Gmaas can tur ...ich has around 73,000 bytes. EDIT: Almighty Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Almighty Gmaas's article has 89
  85 KB (13,971 words) - 18:02, 16 May 2024
• 2020 AIME II Problems/Problem 2
  ...vertices at <math>(0,0), (1,0), (1,1)</math>, and <math>(0,1)</math>. The probability that the slope of the line determined by <math>P</math> and the point <math ...8}.</cmath>Because the entire square has area <math>1,</math> the required probability is <math>\frac{43}{128}</math>. The requested sum is <math>43+128 = 171</ma
  3 KB (403 words) - 17:24, 24 June 2020
• 2021 AIME I Problems/Problem 12
  ...rogs meet. Because the <math>3</math> frogs cannot meet at one vertex, the probability that those two specific frogs meet is <math>\tfrac13</math>. If the expecte Let <math>p_{ij} = P(X_{n+1} = j | X_n = i)</math>, the probability that state <math>i</math> transits to state <math>j</math> on the next step
  8 KB (1,309 words) - 11:57, 3 December 2023
• 2020 CIME I Problems/Problem 13
  ...or <math>\times</math> between every two adjacent numbers, each with equal probability. The expected value of the expression he writes can be expressed as <math>\ [[Category:Intermediate Combinatorics Problems]]
  648 bytes (99 words) - 14:51, 22 December 2023
• 2022 AIME I Problems/Problem 9
  is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is <math>\frac{m}{n},</math> where <math>m</ma ...is is out of <math>\frac{12!}{(2!)^6}</math> possible arrangements, so the probability is: <cmath>\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8
  3 KB (481 words) - 19:31, 31 January 2024
• 2022 AIME I Problems/Problem 12
  ...r each of the <math>n</math> elements, there is a <math>\frac{k}{n}</math> probability that the element will be chosen. To find the sum over all such values, we m [[Category:Intermediate Combinatorics Problems]]
  9 KB (1,471 words) - 16:41, 1 February 2024
• Christmas Invitational Mathematics Examination
  ...ber theory]], and [[probability]] and other secondary school math topics. Problems usually require either very creative use of secondary school curriculum, or ...ems and Solutions]] -- A community effort to provide solutions to all AIME problems from which students can learn.
  3 KB (416 words) - 18:13, 31 August 2020
• 2020 CIME II Problems/Problem 4
  The probability a randomly chosen positive integer <math>N<1000</math> has more digits when ...math> \sum^{3}_{k=1}[8^k-7^k]=8-7+64-49+512-343=185.</cmath> The requested probability is <cmath>\frac{185}{999}=\frac{5}{27},</cmath> and so the answer is <math>
  2 KB (242 words) - 00:49, 19 December 2020
• 2021 AMC 12A Problems/Problem 23
  ...e|[[2021 AMC 10A Problems/Problem 23|2021 AMC 10A #23]] and [[2021 AMC 12A Problems/Problem 23|2021 AMC 12A #23]]}} ...at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
  17 KB (2,801 words) - 07:29, 4 November 2022

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