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• 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

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