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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


Answer: is 489.

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


\(10 \choose 3\) sets of 3 points which form triangles,

fourth distinct segment excluding 3 segments of triangles=45-3=42

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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AMC 8 Math Olympiad USA Math Olympiad

Probability in Divisibility | AMC-10A, 2003 | Problem 15

Try this beautiful problem from AMC 10A, 2003 based on Probability in Divisibility.

Probability in Divisibility – AMC-10A, 2003- Problem 15


What is the probability that an integer in the set \({1,2,3,…,100}\) is divisible by \(2\) and not divisible by \(3\)?

  • \(\frac {33}{100}\)
  • \(\frac{1}{6}\)
  • \(\frac{17}{50}\)
  • \(\frac{1}{2}\)
  • \(\frac{18}{25}\)

Key Concepts


Number system

Probability

divisibility

Check the Answer


Answer: \(\frac{17}{50}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints


There are total number of integers are \(100\).and numer of integers divisible by \(2\) is \(\frac{100}{2}\)=\(50\). Now we have to find out divisible by \(2\) and not divisible by \(3\). so at first we have to find out the numbers of integers which are divisible by \(2\) and \(3\) both…….

can you finish the problem……..

To be divisible by both \(2\) and \(3\), a number must be divisible by the lcm of \((2,3)=6\).

Therefore numbers of integers which are divisible by \(6\)=\(\frac{100}{6}=16\) (between \(1\) & \( 100\))

can you finish the problem……..

Therefore the number of integers which are divisible by \(2\) and not divisible by \(3\)= \(50 – 16=34\).

So require probability= \(\frac{34}{100}=\frac{17}{50}\)

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AMC 10 Math Olympiad Probability USA Math Olympiad

Probability in Game | AMC-10A, 2005 | Problem 18

Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.

Probability in Game – AMC-10A, 2005- Problem 18


Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

  • \(\frac{1}{4}\)
  • \(\frac{1}{6}\)
  • \(\frac{1}{5}\)
  • \(\frac{2}{3}\)
  • \(\frac{1}{3}\)

Key Concepts


Probability

combinatorics

Check the Answer


Answer: \(\frac{1}{5}\)

AMC-10A (2005) Problem 18

Pre College Mathematics

Try with Hints


Given that  The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=\(5\). Now we have to find out the possible order of wins…..

Can you now finish the problem ……….

Possible cases :

If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn’t played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.


According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is \(\frac{ 1}{5}\)

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AMC 10 Math Olympiad USA Math Olympiad

Dice Problem | AMC-10A, 2011 | Problem 14

Try this beautiful problem from Probability based on dice Problem

Dice Problem- AMC-10A, 2011- Problem 14


A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?

  • \(\frac{1}{12}\)
  • \(\frac{7}{12}\)
  • \(\frac{5}{12}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{9}\)

Key Concepts


Probability

dice

circle

Check the Answer


Answer: \(\frac{1}{12}\)

AMC-10A (2011) Problem 14

Pre College Mathematics

Try with Hints


Given that A pair of standard 6-sided fair dice are rolled once. The sum of the numbers rolled determines the diameter of a circle. The numerical value of the area of the circle is less than the numerical value of the circle’s circumference. Let the radius of the circle is \(r\). Then the area of the circle be \(\pi(r)^2\) and  circumference be \(2\pi r\)

can you finish the problem……..

Now according to the given condition we say that \(\pi(r)^2 <2\pi{r}\)\(\Rightarrow r<2\)

As The sum of the numbers rolled determines the diameter of a circle, therefore $r<2$ then the dice must show \((1,1)\),\((1,2)\),\((2,1)\)

can you finish the problem……..

Therefore there $3$ choices out of a total possible of $6 \times 6 =36$, so the probability is \(\frac{3}{36}=\frac{1}{12}\)

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Repeatedly Flipping a Fair Coin | AIME I, 1995| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 repeatedly flipping a fair coin.

Flipping a Fair Coin – AIME I, 1995


Let p be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before on encounters a run of 2 tails. Given that p can be written in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 37
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Probability

Algebra

Check the Answer


Answer: is 37.

AIME I, 1995, Question 15

Elementary Number Theory by David Burton

Try with Hints


Let A be head flipped

B be tail flipped

outcomes are AAAAA, BAAAAA, BB. ABB, AABB, AAABB, AAAABB

with probabilities \(\frac{1}{32}\), \(\frac{1}{64}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\), \(\frac{1}{64}\)

with five heads AAAAA, BAAAAA sum =\(\frac{3}{64}\) and sum of outcomes=\(\frac{34}{64}\)

or, m=3, n=34

or, m+n=37.

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AMC 8 Math Olympiad USA Math Olympiad

Quadratic equation Problem | AMC-10A, 2002 | Problem 12

Try this beautiful problem from Algebra based Quadratic equation.

Quadratic equation Problem – AMC-10A, 2002- Problem 12


Both roots of the quadratic equation \(x^2 – 63x + k = 0\) are prime numbers. The number of possible values of \(k\) is

  • \(0\)
  • \(1\)
  • \(2\)
  • \(4\)
  • more than \(4\)

Key Concepts


Algebra

Quadratic equation

prime numbers

Check the Answer


Answer: \(1\)

AMC-10A (2002) Problem 12

Pre College Mathematics

Try with Hints


The given equation is \(x^2 – 63x + k = 0\). Say that the roots are primes…

Comparing the equation with \(ax^2 +bx+c=0\) we get \(a=1 , b=-63 , c=k\).. Let \(m_1\) & \(m_2 \) be the roots of the given equation…

using vieta’s Formula we may sat that…\(m_1 + m_2 =-(- 63)=63\) and \(m_1 m_2 = k\)

can you finish the problem……..

Now the roots are prime. Sum of the two roots are \(63\) and product is \(k\)

Therefore one root must be \(2\) ,otherwise the sum would be even number

can you finish the problem……..

So other root will be \(63-2\)=\(61\). Therefore product must be \(m_1m_2=122\)

Hence the answer is \(1\)

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AMC 8 Math Olympiad USA Math Olympiad

Largest possible value | AMC-10A, 2004 | Problem 15

Try this beautiful problem from Number system: largest possible value

Largest Possible Value – AMC-10A, 2004- Problem 15


Given that \( -4 \leq x \leq -2\) and \(2 \leq y \leq 4\), what is the largest possible value of \(\frac{x+y}{2}\)

  • \(\frac {-1}{2}\)
  • \(\frac{1}{6}\)
  • \(\frac{1}{2}\)
  • \(\frac{1}{4}\)
  • \(\frac{1}{9}\)

Key Concepts


Number system

Inequality

divisibility

Check the Answer


Answer: \(\frac{1}{2}\)

AMC-10A (2003) Problem 15

Pre College Mathematics

Try with Hints


The given expression is \(\frac{x+y}{x}=1+\frac{y}{x}\)

Now \(-4 \leq x \leq -2\) and \(2 \leq y \leq 4\) so we can say that \(\frac{y}{x} \leq 0\)

can you finish the problem……..

Therefore, the expression \(1+\frac{y}x\) will be maximized when \(\frac{y}{x}\) is minimized, which occurs when \(|x|\) is the largest and \(|y|\) is the smallest.

can you finish the problem……..

Therefore in the region \((-4,2)\) , \(\frac{x+y}{x}=1-\frac{1}{2}=\frac{1}{2}\)

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AMC 8 Math Olympiad USA Math Olympiad

Probability | AMC-10A, 2003 | Problem 8

Try this beautiful problem from Probability based on Positive factors

Probability – AMC-10A, 2003- Problem 8


What is the probability that a randomly drawn positive factor of \(60\) is less than \(7\)?

  • \(\frac{1}{3}\)
  • \(\frac{1}{2}\)
  • \(\frac{3}{4}\)

Key Concepts


Probability

Factors

combinatorics

Check the Answer


Answer: \(\frac{1}{2}\)

AMC-10A (2003) Problem 8

Pre College Mathematics

Try with Hints


Now at first we find out the positive factors of \(60\) are \(1,2,3,4,5,6,10,12,15,20,30,60\).but the positive factors which are less than \(7\) are \(1,2,3,4,5,6\)

Can you now finish the problem ……….

so we may say that any For a positive number \(n\) which is not a perfect square, exactly half of the positive factors will be less than \(\sqrt{n}\).here \(60\) is not a perfect square and \(\sqrt 60 \approx 7.746\).Therefore half of the positive factors will be less than \(7\)

can you finish the problem……..

Therefore the required probability=\(\frac{1}{2}\)

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AMC 8 Math Olympiad USA Math Olympiad

Probability Dice Problem | AMC-10A, 2009 | Problem 22

Try this beautiful problem from Probability based on dice

Probability Dice Problem – AMC-10A, 2009- Problem 22


Two cubical dice each have removable numbers \(1\) through \(6\). The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is \(7\)?

  • \(\frac{3}{8}\)
  • \(\frac{2}{11}\)
  • \(\frac{2}{3}\)
  • \(\frac{1}{3}\)
  • \(\frac{2}{9}\)

Key Concepts


Probability

Combinatorics

Cube

Check the Answer


Answer: \(\frac{2}{11}\)

AMC-10A (2009) Problem 22

Pre College Mathematics

Try with Hints


We assume that the colours of the numbers are different.there are two dices and each of them 1 to 6.after throw,the probability of getting some pair of colors is the same for any two colors.

Therefore there are \(\ 12 \choose 2\)=\(66\) ways to pick to of the colours…

can you finish the problem……..

Now given condition is that the sum will be \(7\).So \(7\) can be obtained by \(1 +6\),\(2+5\),\(3+4\) and  Each number in the bag has two different colors, Therefore each of these three options corresponds to four pairs of colors.SO \(7\) comes from \(3.4\)=\(12\) pairs…..

can you finish the problem……..

So our required probability will be \(\frac{12}{66}=\frac{2}{11}\).

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AMC 8 Math Olympiad USA Math Olympiad

Probability in Coordinates | AMC-10A, 2003 | Problem 12

Try this beautiful problem from Probability based on Coordinates.

Probability in Coordinates – AMC-10A, 2003- Problem 12


A point \((x,y)\) is randomly picked from inside the rectangle with vertices \((0,0)\), \((4,0)\), \((4,1)\), and \((0,1)\). What is the probability that \(x<y\)?

  • \(\frac{1}{8}\)
  • \(\frac{1}{6}\)
  • \(\frac{2}{3}\)

Key Concepts


Number system

adition

Cube

Check the Answer


Answer: \(\frac{1}{8}\)

AMC-10A (2003) Problem 12

Pre College Mathematics

Try with Hints


Probability in Coordinates

The given vertices are \((0,0)\), \((4,0)\), \((4,1)\), and \((0,1)\).if we draw a figure using the given points then we will get a rectangle as shown above.Clearly lengtht of \(OC\)= \(4\) and length of \(AO\)=\(1\).Therefore area of the rectangle is \(4 \times 1=4\).now we have to find out the probability that \(x<y\).so we draw a line \(x=y\) intersects the rectangle at \((0,0)\) and \((1,1)\).can you find out the area with the condition \(x<y\)?

can you finish the problem……..

Coordinate Geometry

Now the line \(x=y\) intersects the rectangle at \((0,0)\) and \((1,1)\).Therefore it will form a Triangle \(\triangle AOD\) (as shown above) whose \(AO=1\) and \(AD=1\).Therefore area of \(\triangle AOD=\frac{1}{2}\) i.e (red region).Now can you find out the probability with the condition \(x<y\)?

can you finish the problem……..

Therefore the required probability (\(x<y\)) is \(\frac{\frac{1}{2}}{4}\)=\(\frac{1}{8}\)

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