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• 2006 AIME II Problems/Problem 10
  ...f the tournament, team <math> A </math> beats team <math> B. </math> The [[probability]] that team <math> A </math> finishes with more points than team <math> B < ...r than <math>A</math>. We let this probability be <math>p</math>; then the probability that <math>A</math> and <math>B</math> end with the same score in these fiv
  6 KB (983 words) - 13:42, 8 December 2021
• 2004 AMC 12B Problems/Problem 23
  ...250,500</math> is a multiple of <math>2004</math>, so there is a very high probability that it is the correct answer. [[Category:Intermediate Algebra Problems]]
  3 KB (533 words) - 14:52, 29 October 2023
• 2005 Alabama ARML TST Problems/Problem 15
  ...positive integers in such a way that when her pair of dice is rolled, the probability of any particular sum occurring is the same as when Virginia rolls her dice Since the probability of any particular sum occurring on Montana's dice is the same as when Virgi
  1 KB (264 words) - 19:43, 19 June 2008
• 2008 AIME I Problems/Problem 9
  ...n of each crate is chosen at random. Let <math>\frac {m}{n}</math> be the probability that the stack of crates is exactly <math>41\mathrm{ft}</math> tall, where ...e heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
  6 KB (909 words) - 15:39, 8 August 2022
• Mock AIME 1 2007-2008 Problems/Problem 8
  ...f ten <math>0</math>s and/or <math>1</math>s is randomly generated. If the probability that the sequence does not contain two consecutive <math>1</math>s can be w ...eed <math>a_n = F_{n+2}</math>, so <math>a_{10} = F_{12} = 144</math>. The probability is <math>\frac{144}{2^{10}} = \frac{9}{64}</math>, and <math>m+n=\boxed{073
  2 KB (260 words) - 18:14, 2 April 2008
• 2008 Mock ARML 1 Problems/Problem 4
  ...rom the hat and replacing it with one of the opposite color. Compute the [[probability]] that, after a sequence of turns, there are <math>5</math> black balls in ...lity of ending with <math>4</math> black balls and a <math>\frac 35</math> probability of ending with <math>2</math> balls. Thus, we have the recursions
  2 KB (242 words) - 09:03, 1 June 2008
• How should I prepare?
  ...your current abilities, you will want to start out with different practice problems, different books, and in different areas of the forums. You should also try ..., Intermediate, and Advanced. The Advanced series, as well as part of the Intermediate series, has not yet been published. These books are indexed [https://artof
  13 KB (1,926 words) - 11:22, 30 November 2023
• 1999 AHSME Problems/Problem 29
  ...<math>P</math> is selected at random inside the circumscribed sphere. The probability that <math>P</math> lies inside one of the five small spheres is closest to Therefore, the probability that <math>P</math> lies inside one of the five small spheres is <math>\fra
  3 KB (522 words) - 11:39, 3 October 2023
• 2010 AIME I Problems/Problem 4
  ...and then Phil flips the three coins. Let <math>\frac {m}{n}</math> be the probability that Jackie gets the same number of heads as Phil, where <math>m</math> and The probability is then <math> \frac{4^2 + 11^2 + 10^2 + 3^2}{28^2} = \frac{246}{784} = \f
  3 KB (470 words) - 22:15, 27 August 2023
• 2010 AIME II Problems/Problem 4
  ...ture gate has been changed to a different gate, again at random. Let the [[probability]] that Dave walks <math>400</math> feet or less to the new gate be a fracti ...cmath>2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76</cmath> so the probability is <math>\frac{76}{132} = \frac{19}{33}</math>. The answer is <math>19 + 33
  2 KB (382 words) - 17:03, 9 August 2018
• 2010 AIME II Problems/Problem 13
  ...th higher numbered cards form another team. Let <math>p(a)</math> be the [[probability]] that Alex and Dylan are on the same team, given that Alex picks one of th [[Category:Intermediate Combinatorics Problems]]
  2 KB (328 words) - 11:15, 7 January 2021
• 2011 AMC 10A Problems/Problem 22
  ...th>E</math> and <math>D</math> have are less than 4.5. (There is a greater probability that <math>A</math>, <math>B</math>, and <math>B</math>, <math>C</math> are [[Category:Intermediate Combinatorics Problems]]
  14 KB (2,425 words) - 09:13, 5 November 2023
• 2011 AIME II Problems/Problem 12
  ...randomly select chairs at a round table that seats nine people. Let the [[probability]] that each delegate sits next to at least one delegate from another countr Use complementary probability and [[Principle of Inclusion-Exclusion]]. If we consider the delegates from
  5 KB (848 words) - 19:15, 30 April 2023
• 2011 AIME II Problems/Problem 15
  ...th> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> i ...(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,</cmath>which means the probability is <cmath>\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.</cmath>The request
  8 KB (1,273 words) - 14:03, 7 January 2023
• 1989 AHSME Problems/Problem 20
  ...een 100 and 200. If <math>\lfloor {\sqrt{x}} \rfloor = 12</math>, find the probability that <math>\lfloor {\sqrt{100x}} \rfloor = 120</math>. (<math>\lfloor {v} \ ...long segment, while the total possibilities region is 25 wide. Thus, the probability is
  1 KB (155 words) - 07:58, 22 October 2014
• Alcumus
  ...nerates high-quality solutions for the students to learn and to do similar problems easily. * Counting and [[Probability]]
  3 KB (533 words) - 10:55, 7 February 2023
• 1989 AHSME Problems/Problem 30
  ...ability is also <math>\frac{7\cdot 13}{20\cdot 19}</math>. Thus, the total probability of the two people being one boy and one girl is <math>\frac{91}{190}</math> [[Category: Intermediate Combinatorics Problems]]
  2 KB (293 words) - 17:13, 24 August 2020
• 2013 AIME I Problems/Problem 6
  ...ooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written ...ll be placed into the same box is <math>126+504+1260=1890</math>. So, the probability of this occurring is <math>\frac{(9\cdot7)(2+8+(4\cdot5))}{12\cdot11\cdot10
  5 KB (831 words) - 17:20, 9 January 2024
• 2014 AIME II Problems/Problem 2
  ...e risk factors, given that he has A and B is <math>\frac{1}{3}</math>. The probability that a man has none of the three risk factors given that he does not have r ...ath>x</math> be the number of men with all three risk factors. Since "the probability that a randomly selected man has all three risk factors, given that he has
  2 KB (371 words) - 15:19, 28 February 2022
• 2014 AIME II Problems/Problem 6
  ...d rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is <math>\frac{p}{q}</math>, where < ...bility that he is using the biased die is <math>\frac{16}{17}</math>. The probability of rolling a third six is
  2 KB (356 words) - 09:03, 14 June 2021

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