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- * [{{SERVER}}/store/item/intro-counting Introduction to Counting & Probability] * [{{SERVER}}/store/item/intermediate-algebra Intermediate Algebra]5 KB (624 words) - 14:58, 13 January 2025
- * 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.23 KB (3,038 words) - 18:33, 15 February 2025
- * 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 (902 words) - 14:34, 13 January 2025
- ...use during the test; however, calculators were never required to solve any problems, and students who did not use calculators were not disadvantaged. ...mber theory]], and [[probability]] and other secondary school math topics. Problems are designed to be solvable by students without any background in calculus.5 KB (646 words) - 03:52, 3 February 2025
- ...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, or5 KB (669 words) - 03:52, 3 February 2025
- In contest problems, Fermat's Little Theorem is often used in conjunction with the [[Chinese Re ....khanacademy.org/computing/computer-science/cryptography/random-algorithms-probability/v/fermat-s-little-theorem-visualization Video explanation]15 KB (2,618 words) - 11:03, 19 February 2025
- ...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 (590 words) - 10:52, 28 September 2024
- ...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) - 16:20, 16 June 2023
- ==Problems== ...uad \mathrm{(E) \ 9 } </math><div style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div>7 KB (1,111 words) - 13:57, 24 June 2024
- ...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 measu2 KB (343 words) - 14:39, 14 June 2023
- ...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 multiply this number by the probability that each case occurs.10 KB (1,840 words) - 14:01, 4 July 2024
- ...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 rol4 KB (628 words) - 10:28, 14 April 2024
- ...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 <math5 KB (836 words) - 06:53, 15 October 2023
- ...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 = \boxed8 KB (1,283 words) - 18:19, 8 May 2024
- ..., 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}{12 KB (330 words) - 12:42, 1 January 2015
- ...- 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 subtracti9 KB (1,458 words) - 21:34, 2 February 2025
- ...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) - 19:26, 14 January 2023
- ...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 multiply7 KB (1,115 words) - 23:52, 6 September 2023
- ...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>19 KB (3,128 words) - 20:38, 23 July 2024
- ...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 position8 KB (1,269 words) - 09:55, 26 June 2024