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• Art of Problem Solving
  ...tofproblemsolving.com/store/item/intro-counting Introduction to Counting & Probability] * [https://artofproblemsolving.com/store/item/intermediate-algebra Intermediate Algebra]
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• Math books
  * Intermediate is recommended for students who can expect to pass the AMC 10/12. ...combinatorics, and number theory, along with sets of accompanying practice problems at the end of every section.
  24 KB (3,177 words) - 12:53, 20 February 2024
• Math textbooks
  * Intermediate is recommended for students grades 9 to 12. ...combinatorics, and number theory, along with sets of accompanying practice problems at the end of every section.
  7 KB (901 words) - 14:11, 6 January 2022
• AMC 12
  ...use during the test; however, calculators were never required to solve any problems, and students who did not use calculators were not disadvantaged. ...ber theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.
  4 KB (520 words) - 12:11, 13 March 2024
• American Invitational Mathematics Examination
  ...correct answers receive one point of credit, making the maximum score 15. Problems generally increase in difficulty as the exam progresses - the first few que ...ber theory]], and [[probability]] and other secondary school math topics. Problems usually require either very creative use of secondary school curriculum, or
  8 KB (1,057 words) - 12:02, 25 February 2024
• Probability
  ...Examples include the [[Monty Hall paradox]] and the [[birthday problem]]. Probability can be loosely defined as the chance that an event will happen. == Introductory Probability ==
  4 KB (588 words) - 12:47, 2 October 2022
• Complementary counting
  ...don't want, then subtracts that from the total number of possibilities. In problems that involve complex or tedious [[casework]], complementary counting is oft == Complementary Probability ==
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• Fibonacci sequence
  ==Problems== ...uad \mathrm{(E) \ 9 } </math><div style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div>
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• 2006 AMC 12A Problems/Problem 22
  ...gular polygon | regular]] [[hexagon]] of side length <math>2</math>. The [[probability]] that three entire sides of hexagon are visible from a randomly chosen poi ...y viewed. Since there are six such pairs of sides, there are six arcs. The probability of choosing a point is <math>1 / 2</math>, or if the total arc degree measu
  2 KB (343 words) - 15:39, 14 June 2023
• 2005 AMC 12B Problems/Problem 25
  ...f the four adjacent vertices, each with equal [[probability]]. What is the probability that no two ants arrive at the same vertex? ...ants can do their desired migration, and then multiple this number by the probability that each case occurs.
  10 KB (1,840 words) - 21:35, 7 September 2023
• 2005 AIME II Problems/Problem 2
  ...randomly put three rolls in a bag for each of the guests. Given that the [[probability]] each guest got one roll of each type is <math> \frac mn, </math> where <m Use [[construction]]. We need only calculate the probability the first and second person all get a roll of each type, since then the rol
  4 KB (628 words) - 11:28, 14 April 2024
• 2004 AIME I Problems/Problem 10
  ...so that the circle lies completely within the rectangle. Given that the [[probability]] that the circle will not touch diagonal <math> AC </math> is <math> m/n, The probability is <math>\frac{2[A'B'C']}{34 \times 13} = \frac{375}{442}</math>, and <math
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME II Problems/Problem 10
  ...h>1</math>'s. If a number is chosen at random from <math> S, </math> the [[probability]] that it is divisible by <math>9</math> is <math> p/q, </math> where <math The probability is <math>\frac{133}{780}</math>, and the answer is <math>133 + 780 = \boxed
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 2
  ..., then Mary picks two of the remaining candies at random. Given that the [[probability]] that they get the same color combination, irrespective of order, is <math ...andies is <math>\frac{7\cdot8}{18\cdot17} = \frac{28}{153}</math>. So the probability that they both pick two red candies is <math>\frac{9}{38} \cdot \frac{28}{1
  2 KB (330 words) - 13:42, 1 January 2015
• 1983 AIME Problems/Problem 7
  ...- and are sent off to slay a troublesome dragon. Let <math>P</math> be the probability that at least two of the three had been sitting next to each other. If <mat We can use [[complementary counting]], by finding the probability that none of the three knights are sitting next to each other and subtracti
  9 KB (1,392 words) - 20:37, 19 January 2024
• 1983 AIME Problems/Problem 10
  ...g at the sequences which have one digit repeated twice, we notice that the probability that the sequence starts with 1 is <math>\frac{1}{10}</math>. This means we * [[AIME Problems and Solutions]]
  5 KB (855 words) - 20:26, 14 January 2023
• 1984 AIME Problems/Problem 11
  ...being equally likely. Let <math>\frac m n</math> in lowest terms be the [[probability]] that no two birch trees are next to one another. Find <math>m+n</math>. ...ave them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply
  7 KB (1,115 words) - 00:52, 7 September 2023
• 1985 AIME Problems/Problem 12
  ...the vertex at its opposite end. Let <math>p = \frac{n}{729}</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math> For all nonnegative integers <math>k,</math> let <math>P(k)</math> be the probability that the bug is at vertex <math>A</math> when it has crawled exactly <math>
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1987 AIME Problems/Problem 13
  ...and are in random order. Let <math>p/q</math>, in lowest terms, be the [[probability]] that the number that begins as <math>r_{20}</math> will end up, after one Thus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position
  3 KB (514 words) - 21:27, 31 December 2023
• 1988 AIME Problems/Problem 15
  This problem is very similar to AoPS Intermediate Counting and Probability problem 4.8(which is another AIME problem but I don't know which one) [[Category:Intermediate Combinatorics Problems]]
  7 KB (1,186 words) - 10:16, 4 June 2023
• 1988 AIME Problems/Problem 5
  Let <math>m/n</math>, in lowest terms, be the [[probability]] that a randomly chosen positive [[divisor]] of <math>10^{99}</math> is an ...2^{11}5^{11}</math> has, which is <math>(11 + 1)(11 + 1) = 144</math>. Our probability is <math>\frac{m}{n} = \frac{144}{10000} = \frac{9}{625}</math>, and <math>
  822 bytes (108 words) - 22:21, 6 November 2016
• 1989 AIME Problems/Problem 5
  ...ng heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based
  2 KB (258 words) - 00:07, 25 June 2023
• 1990 AIME Problems/Problem 9
  ...math> times. Let <math>\frac{i}{j}^{}_{}</math>, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math> ...of <math>2^{10}</math> possible flips of <math>10</math> coins, making the probability <math>\frac{144}{1024} = \frac{9}{64}</math>. Thus, our solution is <math>9
  3 KB (425 words) - 12:36, 12 May 2024
• 1991 AIME Problems/Problem 13
  ...that, when two socks are selected randomly without replacement, there is a probability of exactly <math>\frac{1}{2}</math> that both are red or both are blue. Wha ...ber of red and blue socks, respectively. Also, let <math>t=r+b</math>. The probability <math>P</math> that when two socks are drawn randomly, without replacement,
  7 KB (1,328 words) - 20:24, 5 February 2024
• 1991 AIME Problems/Problem 10
  ...ed when <math>bbb^{}_{}</math> is transmitted. Let <math>p</math> be the [[probability]] that <math>S_a^{}</math> comes before <math>S_b^{}</math> in alphabetical The probability is <math>p=\sum P_a \cdot (27 - S_b)</math>, so the answer turns out to be
  5 KB (813 words) - 06:10, 25 February 2024
• 1993 AIME Problems/Problem 7
  ...m the remaining set of <math>997</math> numbers. Let <math>p</math> be the probability that, after suitable rotation, a brick of dimensions <math>a_1 \times a_2 \ ...6)\cdot5\cdot6^2</math> valid ordered <math>6</math>-tuples. The requested probability is <cmath>p=\frac{C(1000,6)\cdot5\cdot6^2}{P(1000,6)}=\frac{C(1000,6)\cdot5
  5 KB (772 words) - 09:04, 7 January 2022
• 1994 AIME Problems/Problem 9
  ...which match; otherwise the drawing continues until the bag is empty. The probability that the bag will be emptied is <math>p/q,\,</math> where <math>p\,</math> Let <math>P_k</math> be the [[probability]] of emptying the bag when it has <math>k</math> pairs in it. Let's conside
  3 KB (589 words) - 14:18, 21 July 2019
• 1995 AIME Problems/Problem 15
  Let <math>p_{}</math> be the [[probability]] that, in the process of repeatedly flipping a fair coin, one will encount ...T</tt> or the sequence could start with a block of <tt>H</tt>'s, the total probability is that <math>3/2</math> of it has to start with an <tt>H</tt>.
  6 KB (979 words) - 13:20, 11 April 2022
• 1995 AIME Problems/Problem 3
  ...t, right, up, or down, all four equally likely. Let <math>p</math> be the probability that the object reaches <math>(2,2)</math> in six or fewer steps. Given th ...>\frac{4!}{2!2!} = 6</math> ways for these four steps of occuring, and the probability is <math>\frac{6}{4^{4}}</math>.
  3 KB (602 words) - 23:15, 16 June 2019
• 1997 AIME Problems/Problem 14
  ...equation <math>z^{1997}-1=0</math>. Let <math>\frac{m}{n}</math> be the [[probability]] that <math>\sqrt{2+\sqrt{3}}\le\left|v+w\right|</math>, where <math>m</ma ...<math>n</math> can have <math>1996</math> possible values. Therefore, the probability is <math>\frac{332}{1996}=\frac{83}{499}</math>. The answer is then <math>
  5 KB (874 words) - 22:30, 1 April 2022
• 1998 AIME Problems/Problem 9
  ...en 9 a.m. and 10 a.m., and stay for exactly <math>m</math> minutes. The [[probability]] that either one arrives while the other is in the cafeteria is <math>40 \ ...rea of the unshaded region over the area of the total region, which is the probability that the mathematicians do not meet:
  4 KB (624 words) - 18:34, 18 February 2018
• 1998 AIME Problems/Problem 4
  ...selects and keeps three of the tiles, and sums those three values. The [[probability]] that all three players obtain an [[odd]] sum is <math>m/n,</math> where < In order to calculate the probability, we need to know the total number of possible distributions for the tiles.
  5 KB (917 words) - 02:37, 12 December 2022
• 1999 AIME Problems/Problem 13
  ...team has a <math>50 \%</math> chance of winning any game it plays. The [[probability]] that no two teams win the same number of games is <math>\frac mn,</math> The desired probability is thus <math>\frac{40!}{2^{780}}</math>. We wish to simplify this into the
  2 KB (329 words) - 01:38, 6 October 2015
• 2000 AIME I Problems/Problem 5
  ...[[probability]] that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and < ...rinciple of Inclusion-Exclusion]] still requires us to find the individual probability of each box.
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2001 AIME I Problems/Problem 15
  ...gular [[octahedron]] so that each face contains a different number. The [[probability]] that no two consecutive numbers, where <math>8</math> and <math>1</math> ...rrespond to octagonal circuits formed exclusively from cube diagonals. The probability of randomly choosing such a permutation is <math>\tfrac{480}{8!}=\tfrac{1}{
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 10
  ...math> Two distinct points are randomly chosen from <math>S.</math> The [[probability]] that the [[midpoint]] of the segment they determine also belongs to <math Ignore the distinct points condition. The probability that the midpoint is in <math>S</math> is then
  8 KB (1,187 words) - 02:40, 28 November 2020
• 2001 AIME I Problems/Problem 6
  A fair die is rolled four times. The [[probability]] that each of the final three rolls is at least as large as the roll prece ...2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126</math>. The requested probability is <math>\frac{126}{6^4} = \frac{7}{72}</math> and our answer is <math>\box
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2002 AIME I Problems/Problem 1
  ...ven that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-let ...10^3} = \frac 1{10}</math>. Similarly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome.
  3 KB (369 words) - 23:36, 6 January 2024
• 2003 AIME I Problems/Problem 11
  ...erval]] <math> 0^\circ < x < 90^\circ. </math> Let <math> p </math> be the probability that the numbers <math> \sin^2 x, \cos^2 x, </math> and <math> \sin x \cos ...spect to interchanging <math>\sin</math> and <math>\cos</math>, and so the probability is symmetric around <math>45^\circ</math>. Thus, take <math>0 < x < 45</ma
  2 KB (284 words) - 13:42, 10 October 2020
• 2003 AIME II Problems/Problem 13
  ...ed, and crawls along a side of the triangle to that vertex. Given that the probability that the bug moves to its starting vertex on its tenth move is <math>m/n,</ Let <math>P_n</math> represent the probability that the bug is at its starting vertex after <math>n</math> moves. If the b
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2002 AIME II Problems/Problem 12
  ...the ratio of shots made to shots attempted after <math>n</math> shots. The probability that <math>a_{10} = .4</math> and <math>a_n\le.4</math> for all <math>n</ma ...ossible sequence occurring is <math>(.4)^4(.6)^6</math>. Hence the desired probability is
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2001 AIME II Problems/Problem 11
  ...Truncator will win, lose, or tie are each <math>\frac {1}{3}</math>. The [[probability]] that Club Truncator will finish the season with more wins than losses is ...Thus, by the [[complement principle]], the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
  3 KB (415 words) - 23:25, 20 February 2023
• 2001 AIME II Problems/Problem 9
  ...or red. For each square, either color is equally likely to be used. The [[probability]] of obtaining a grid that does not have a 2-by-2 red square is <math>\frac ...5=417</math> ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a <math>2 \times 2</math> red square
  8 KB (1,207 words) - 20:04, 5 September 2023
• 2000 AIME II Problems/Problem 3
  ...at these cards are not returned to the deck, let <math>m/n</math> be the [[probability]] that two randomly selected cards also form a pair, where <math>m</math> a [[Category:Intermediate Combinatorics Problems]]
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• Subject Classes
  * Intermediate (at the level of hardest [[AMC 12]] problems, the [[AIME]], [[ARML]], and the [[Mandelbrot Competition]]). * [[Introduction to Counting & Probability Course]] [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#be
  2 KB (303 words) - 16:02, 11 July 2006
• Expected value
  A video that goes over the type of Expected value, practical examples, and problems: https://youtu.be/TCFoRx2R2ew ...that particular outcome. If the event <math>Z</math> has a [[continuous]] probability distribution, then <math>E(Z) = \int_z P(z)\cdot z\ dz</math>.
  5 KB (789 words) - 20:56, 10 May 2024
• 2005 Alabama ARML TST Problems/Problem 1
  ...of the other die are <math>1, 2, 2, 3, 3,\text{ and }4</math>. Find the [[probability]] of rolling a sum of <math>9</math> with these two dice. ..., out of a total of <math>6^2</math> possible two-roll combinations, for a probability of <math>\frac 19</math>.
  1 KB (210 words) - 01:30, 3 January 2023
• ITest
  ...est is varied, ranging from simple arithmetic problems to complex Olympiad problems. Winning teams earn recent technology prizes like a video game console of ...ol math concepts, including [[algebra]], [[geometry]], pre-[[calculus]], [[probability]], and [[logic]]. Graphing calculators are allowed, but computers are not
  2 KB (264 words) - 20:31, 13 December 2018
• Parity
  ...asic, checking the parity of numbers is often an useful tactic for solving problems, especially with [[proof by contradiction]]s and [[casework]]. == Problems ==
  4 KB (694 words) - 22:00, 12 January 2024
• 2006 AIME II Problems/Problem 5
  ...obability of obtaining a sum of 7 is <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math> is <math> m/n, </math> where <math> m ...be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <mat
  5 KB (712 words) - 12:10, 5 November 2023
• 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
• 2014 AIME II Problems/Problem 13
  ...shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer <math>k<5</math>, no collection of <math>k< ...ngs. Therefore, taking these cases out of a total of <math>10!</math>, the probability is <math>\frac{1}{10}+\frac{1}{50} = \frac{3}{25}</math>, for an answer of
  3 KB (509 words) - 15:33, 5 October 2023
• 2014 AIME I Problems/Problem 2
  ...h>N</math> blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is <math>0.58</math>. Find <math>N</m First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to <m
  1 KB (173 words) - 20:12, 15 January 2022
• 1993 AHSME Problems/Problem 24
  ...by one, pennies are drawn at random from the box and not replaced. If the probability is <math>a/b</math> that it will take more than four draws until the third So, probability for Case 1 to occur is:
  3 KB (558 words) - 00:25, 23 December 2020
• 1992 AHSME Problems/Problem 29
  ...urning up heads. If this coin is tossed <math>50</math> times, what is the probability that the total number of heads is even? Denote <math>P_n</math> as the probability that the number of heads is even. Then we have two cases:
  3 KB (472 words) - 19:29, 20 June 2021
• 1970 AHSME Problems/Problem 33
  ...>0000-9999</math>. We have <math>40000</math> digits and each has equal an probability of being <math>0,1,2....9</math>. [[Category: Intermediate Algebra Problems]]
  3 KB (418 words) - 15:54, 15 December 2018
• 1970 AHSME Problems/Problem 31
  ...e-digit numbers in which the sum of the digits is equal to 43, what is the probability that this number will be divisible by 11? [[Category: Intermediate Algebra Problems]]
  1 KB (172 words) - 10:53, 17 April 2022
• 2007 iTest Problems/Problem 19
  ...ail then <math>1</math> head is <math>\frac{1}{4}</math>. From there, the probability of getting <math>n</math> tails then <math>1</math> head is <math>\frac{1}{ [[Category:Intermediate Combinatorics Problems]]
  3 KB (397 words) - 01:18, 29 October 2023
• 2007 iTest Problems/Problem 21
  What is the probability that the value of the expression James wrote down is <math>7</math>? ...signs and minus signs between consecutive ones is <math>2^{14}</math>, the probability of getting a <math>7</math> is <math>\boxed{\textbf{(L) } \frac{1001}{2^{14
  2 KB (247 words) - 03:34, 14 June 2018
• 2007 iTest Problems/Problem 36
  ...n, m and n are two relatively prime positive integers such that m/n is the probability that the equation <math>x^4+25b^2=(4b^2-10b)x^2</math> has <math>\textit{at ...fore, in interval notation, <math>b \in [-17,0) \cup [5,17]</math>, so the probability that the equation has at least two distinct real solutions when <math>b</ma
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• 2007 iTest Problems/Problem 42
  ...ming the stuntman is equally likely to land on any point in the field, the probability that he lands closer to the fence than to the flag pole can be written in s ...etry]], the square can be divided into four look-alike parts. To find the probability of landing closer to the fence, subtract the outlined red area from the are
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• 2007 iTest Problems/Problem 44
  ...> and <math>b</math> are natural numbers such that <math>a/b</math> is the probability that <math>N</math> and <math>n^3-36n</math> are relatively prime, find the ...nce a number is picked at random from <math>2007^{2007}</math> values, the probability that <math>N</math> and <math>n^3 - 36n</math> are relatively prime is <mat
  2 KB (305 words) - 23:24, 14 June 2018
• 2007 iTest Problems/Problem 59
  ...y many games (eh, they're in Physicist Heaven, we can bend the rules), the probability that they are ever tied in total wins after they start (they have the same ...number of Fermi wins followed by the number of Feynman wins, and call the probability of Fermi winning <math>p</math>.
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• Mock AIME 5 2005-2006 Problems/Problem 7
  ...also lands completely within one of the circles. Let <math>P</math> be the probability that, when flipped onto the grid, the coin lands completely within one of t [[Category:Intermediate Number Theory Problems]]
  853 bytes (134 words) - 21:18, 8 October 2014
• 2012 UNCO Math Contest II Problems/Problem 8
  The probability that a six is rolled on any given roll is <math>1/6</math>, so on average i [[Category:Intermediate Combinatorics Problems]]
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• 2014 UNCO Math Contest II Problems/Problem 8
  ...Alice makes the first hit and alternates hits with the Queen, what is the probability that Alice Now what is the probability that Alice is the first player to hit the goal post with the ball?
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• 2014 UNCO Math Contest II Problems/Problem 9
  In the Queen’s croquet, as described in Problem <math>8</math>, what is the probability that the ball hits the Zero probability for <math>n=1</math> and <math>\frac{F_{n-1}}{2^{2n}}</math> for <math>n\ge
  493 bytes (80 words) - 03:32, 13 January 2019
• 2014 UNCO Math Contest II Problems/Problem 10
  ...uests make their guesses at random, perhaps by tossing a coin. What is the probability (b) Tweedledum is the first guest to arrive. What is the probability that one or the other of Tweedledee
  2 KB (297 words) - 03:33, 13 January 2019
• 1988 AHSME Problems/Problem 28
  An unfair coin has probability <math>p</math> of coming up heads on a single toss. Let <math>w</math> be the probability that, in <math>5</math> independent toss of this coin,
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• 1987 AHSME Problems/Problem 26
  <math>2</math> and <math>1</math> in the second. What is the probability that the two integers sum to <math>3</math>? [[Category: Intermediate Algebra Problems]]
  2 KB (235 words) - 13:01, 31 March 2018
• 1986 AHSME Problems/Problem 22
  ...rs are picked at random from <math>\{1,2,3,\ldots,10\}</math>. What is the probability that, among those selected, the second smallest is <math>3</math>? [[Category: Intermediate Combinatorics Problems]]
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• 2015 UNCO Math Contest II Problems/Problem 8
  at a time, without any going back in. What is the probability that the sequence of numbers [[Category:Intermediate Probability Problems]]
  887 bytes (138 words) - 03:35, 13 January 2019
• 2015 AMC 10A Problems/Problem 25
  ...ts are chosen independently at random on the sides of <math>S</math>. The probability that the straight-line distance between the points is at least <math>\dfrac ...lity. (Alternatively, one can set up a coordinate system and use geometric probability.)
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• 2018 UNCO Math Contest II Problems/Problem 7
  positive integers less than or equal to twelve. What is the probability that <math>x</math> is a multiple of By counting the number of solutions, the required probability <math>=\frac{1\times1 + 5\times11}{12\times12} = \frac{56}{144} = \frac{7}{
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• Gmaas
  ...roblem. Therefore, I proved that you cannot use the Games theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Games can turn things ...oblems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes
  69 KB (11,805 words) - 20:49, 18 December 2019
• 2016 UNCO Math Contest II Problems/Problem 5
  probability 10%), be rolled back down into the valley (with probability 50% ), or be split by a thunderbolt into two rocks that are both rolled down into the valley (with probability 40%).
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• 2016 UNCO Math Contest II Problems/Problem 6
  (a) Given the situation in Question <math>5</math>, what is the probability that first morning, what is the probability that the Olympian rocks are never all vaporized?
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• 2017 AMC 12B Problems/Problem 22
  ...ts the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins? ...th> ways for this to happen, along with <math>12^3</math> total cases. The probability we're asking for is thus <math>\frac{45}{(12^3)}= \boxed{\textbf{(B)}\frac{
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• 2017 AMC 12B Problems/Problem 20
  ...and uniformly at random from the interval <math>(0,1)</math>. What is the probability that <math>\lfloor\log_2x\rfloor=\lfloor\log_2y\rfloor</math>? ...g the formula for the sum of an infinite geometric series, we get that the probability is <math>\frac{\frac{1}{4}}{1 - \frac{1}{4}} = \boxed{\textbf{(D)}\frac{1}{
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• 2018 AIME II Problems/Problem 6
  ...hosen randomly and uniformly from the interval <math>[-20, 18]</math>. The probability that the roots of the polynomial [[Category:Intermediate Algebra Problems]]
  2 KB (286 words) - 12:20, 27 March 2022
• 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
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• 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]]
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• 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
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• 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>
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• 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
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• 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.
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• 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>.
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• 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 (13,954 words) - 17:25, 22 March 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
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• 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
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• 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.
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• 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?
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• 2021 AIME II Problems/Problem 8
  ...lected at random so that each of the possible moves is equally likely. The probability that after exactly <math>8</math> moves that ant is at a vertex of the top Finally, the requested probability is <cmath>\frac{N(8,\mathrm{BT})+N(8,\mathrm{TT})}{N(8,\mathrm{BB})+N(8,\ma
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• 2022 AIME II Problems/Problem 2
  ...c34</math>. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is <math>\frac{p}{q}</math>, where <math> ...h> players, and the arrow is from the winner to the loser with the winning probability as the label.
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• 2021 Fall AMC 12B Problems/Problem 23
  ...utive integers out of <math>{30 \choose 2}</math> possible pairs, thus the probability that any given pair is consecutive is <math>\frac{29}{{30 \choose 2}}</math [[Category:Intermediate Combinatorics Problems]]
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• 2021 Fall AMC 12B Problems/Problem 17
  ...grid lines randomly and independently with equal probability. What is the probability that after <math>5</math> moves the bug never will have been more than <mat ...) = 6C(4) + 3V(4) = 6 \cdot 60 + 3 \cdot 192 = 936,</math> and the desired probability is thus <math>\frac{936}{6^5} = \boxed{\textbf{(A)}\ \frac{13}{108}}.</math
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• 2023 AIME I Problems/Problem 6
  ...guess the color with <math>a</math> unrevealed cards, which succeeds with probability <math>\frac{a}{a+b}.</math> There are always <math>3</math> unrevealed cards of each color, so the probability of guessing correctly is <math>\frac{1}{2}</math>.
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• 2023 AIME II Problems/Problem 6
  ...e chosen independently and uniformly at random from inside the region. The probability that the midpoint of <math>\overline{AB}</math> also lies inside this L-sha ...right square. This occurs with <math>\frac{2}{3} \cdot \frac{1}{3}</math> probability.
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• Know the Facts: Gmaas
  ...roblem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Gmaas can turn things ...oblems, which has around 73,000 bytes. EDIT: Gmaas has beaten 2008 most iT Problems! It is the longest AoPS Wiki article ever. Gmaas's article has 89,119 bytes
  88 KB (14,928 words) - 13:54, 29 April 2024
• 2023 AMC 12A Problems/Problem 17
  ...revious jumps, Flora leaps a positive integer distance <math>m</math> with probability <math>\frac{1}{2^m}</math>. What is the probability that Flora will eventually land at 10?
  7 KB (1,072 words) - 10:46, 22 February 2024