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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 ==
  8 KB (1,192 words) - 17:20, 16 June 2023
• 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) - 19:31, 30 July 2021
• 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]]
  1 KB (191 words) - 04:27, 4 November 2022
• 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

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