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  • 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%).
    964 bytes (160 words) - 04:02, 13 January 2019
  • (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?
    974 bytes (151 words) - 04:03, 13 January 2019
  • ...e 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is We let <math>c, e,</math> and <math>m</math> be the probability of reaching a corner before an edge when starting at an "inside corner" (e.
    4 KB (588 words) - 16:19, 17 May 2024
  • ...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{
    12 KB (1,935 words) - 11:37, 16 December 2023
  • ...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}{
    2 KB (327 words) - 17:24, 27 September 2022
  • ...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
  • ...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
    11 KB (1,860 words) - 13:12, 24 January 2024
  • ...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
  • ...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]]
    1 KB (240 words) - 00:03, 6 August 2018
  • ...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
  • ...> 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
    2 KB (284 words) - 23:27, 6 May 2019
  • 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>
    6 KB (950 words) - 14:53, 7 December 2021
  • 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
    3 KB (461 words) - 13:09, 2 January 2020
  • ...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
  • 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.
    19 KB (2,941 words) - 16:16, 7 June 2024
  • ...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>.
    5 KB (819 words) - 16:21, 20 December 2019
  • ...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
  • ...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 (13,971 words) - 18:02, 16 May 2024
  • ...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
    3 KB (403 words) - 17:24, 24 June 2020
  • ...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
    8 KB (1,309 words) - 11:57, 3 December 2023

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