Categories

## Probability in Games | AIME I, 1999 | Question 13

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 1999 based on Probability in games.

## Probability in Games – AIME I, 1999 Question 13

Forty teams play a tournament in which every team plays every other team exactly once. No ties occur,and each team has a 50% chance of winning any game it plays.the probability that no two team win the same number of games is $\frac{m}{n}$ where m and n are relatively prime positive integers, find $log_{2}n$

• 10
• 742
• 30
• 11

### Key Concepts

Probability

Theory of equations

Combinations

AIME, 1999 Q13

Course in Probability Theory by Kai Lai Chung .

## Try with Hints

${40 \choose 2}$=780 pairings with $2^{780}$ outcomes

no two teams win the same number of games=40! required probability =$\frac{40!}{2^{780}}$

the number of powers of 2 in 40!=[$\frac{40}{2}$]+[$\frac{40}{4}$]+[$\frac{40}{8}$]+[$\frac{40}{16}$]+[$\frac{40}{32}$]=20+10+5+2+1=38 then 780-38=742.

Categories

## Probability of tossing a coin | AIME I, 2009 | Question 3

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on Probability of tossing a coin.

## Probability of tossing a coin – AIME I, 2009 Question 3

A coin that comes up heads with probability p>0and tails with probability (1-p)>0 independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac{1}{25}$ the probability of five heads and three tails. Let p=$\frac{m}{n}$ where m and n are relatively prime positive integers. Find m+n.

• 10
• 20
• 30
• 11

### Key Concepts

Probability

Theory of equations

Polynomials

AIME, 2009

Course in Probability Theory by Kai Lai Chung .

## Try with Hints

here $\frac{8!}{3!5!}p^{3}(1-p)^{5}$=$\frac{1}{25}\frac{8!}{5!3!}p^{5}(1-p)^{3}$

then $(1-p)^{2}$=$\frac{1}{25}p^{2}$ then 1-p=$\frac{1}{5}p$

then p=$\frac{5}{6}$ then m+n=11

Categories

## Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

## Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of $2010^{2}$, she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

$(2+1)^{4}$ divisors, $2^{4}$ are squares

probability is $\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}$ implies m+n=107

Categories

## Probability Problem | Combinatorics | AIME I, 2015 – Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Probability.

## Probability Problem – AIME I, 2015

In a drawer Sandy has 5 pairs of socks, each pair a different color. on monday sandy selects two individual socks at random from the 10 socks in the drawer. On tuesday Sandy selects 2 of the remaining 8 socks at random and on wednesday two of the remaining 6 socks at random. The probability that wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 341
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Theory of Equations

Probability

AIME, 2015, Question 5

Geometry Revisited by Coxeter

## Try with Hints

Wednesday case – with restriction , select the pair on wednesday in $5 \choose 1$ ways

Tuesday case – four pair of socks out of which a pair on tuesday where a pair is not allowed where 4 pairs are left,the number of ways in which this can be done is $8 \choose 2$ – 4

Monday case – a total of 6 socks and a pair not picked $6 \choose 2$ -2

by multiplication and principle of combinatorics $\frac{(5)({5\choose 2} -4)({6 \choose 2}-2)}{{10 \choose 2}{8 \choose 2}{6 \choose 2}}$=$\frac{26}{315}$. That is 341.

Categories

## Coordinate Geometry Problem | AIME I, 2009 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

## Coordinate Geometry Problem – AIME 2009

Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

• is 107
• is 600
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Geometry

AIME, 2009, Question 11

Geometry Revisited by Coxeter

## Try with Hints

let P and Q be defined with coordinates; P=($x_1,y_1)$ and Q($x_2,y_2)$. Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=$\frac{1}{2}$OY.OX=$\frac{1}{2}$2009.49 And [OYP]=$\frac{1}{2}$$2009x_1$  and [OXQ]=$\frac{1}{2}$(49)$y_2$.

2009.49 is odd, area OYX not integer of form k+$\frac{1}{2}$ where k is an integer

41x+y=2009 taking both 25  $\frac{25!}{2!23!}+\frac{25!}{2!23!}$=300+300=600.

.

Categories

## Probability Problem | AMC 8, 2016 | Problem no. 21

Try this beautiful problem from Probability.

## Problem based on Probability | AMC-8, 2016 | Problem 21

A box contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?

• $\frac{3}{5}$
• $\frac{2}{5}$
• $\frac{1}{4}$

### Key Concepts

probability

combination

fraction

Answer: $\frac{2}{5}$

AMC-8, 2016 problem 21

Challenges and Thrills in Pre College Mathematics

## Try with Hints

There are 5 Chips, 3 red and 2 green

Can you now finish the problem ……….

We draw the chips boxes in such a way that we do not stop when the last chip of color is drawn. one at a time without replacement

Can you finish the problem……..

There are 5 Chips, 3 red and 2 green

we draw the chips boxes in such a way that we do not stop when the last chip of color is drawn.

if we draw all the green chip boxes then the last box be red or if we draw all red boxes then the last box be green

but we draw randomly. there are 3 red boxes and 2 green boxes

Therefore the probability that the 3 reds are drawn=$\frac{2}{5}$

Categories

## Problem from Probability | AMC 8, 2004 | Problem no. 21

Try this beautiful problem from Probability from AMC 8, 2004.

## Problem from Probability | AMC-8, 2004 | Problem 21

Spinners A and B  are spun. On each spinner, the arrow is equally likely to land on each number. What is the probability that the product of the two spinners’ numbers is even?

• $\frac{2}{3}$
• $\frac{1}{3}$
• $\frac{1}{4}$

### Key Concepts

probability

Equilly likely

Number counting

Answer: $\frac{2}{3}$

AMC-8, 2004 problem 21

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Even number comes from multiplying an even and even, even and odd, or odd and even

Can you now finish the problem ……….

A odd number only comes from multiplying an odd and odd…………..

can you finish the problem……..

We know that even number comes from multiplying an even and even, even and odd, or odd and even

and also a odd number only comes from multiplying an odd and odd,

There are few cases to find the probability of spinning two odd numbers from  1

Multiply the independent probabilities of each spinner getting an odd number together and subtract it from  1 we get…….

$1 – \frac{2}{4} \times \frac{2}{3}$= $1 – \frac{1}{3} = \frac{2}{3}$

Categories

## Probability | AMC 8, 2004 | Problem no. 22

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

## Probability | AMC-8, 2004 |Problem 22

At a party there are only single women and married men with their wives. The probability that a randomly selected woman is single is$\frac{2}{5}$. What fraction of the people in the room are married men?

• $\frac{3}{8}$
• $\frac{1}{2}$
• $\frac{1}{4}$

### Key Concepts

probability

combination

Number counting

Answer: $\frac{3}{8}$

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

## Try with Hints

Find the married men in the room …

Can you now finish the problem ……….

Find the total people

can you finish the problem……..

Assume that there are 10 women in the room, of which $10 \times \frac{2}{5}$=4 are single and 10-4=6 are married. Each married woman came with her husband,

so there are 6 married men in the room

Total man=10+6=16 people

Now The fraction of the people that are married men is $\frac{6}{16}=\frac{3}{8}$

Categories

## Probability Biased and Unbiased | AIME I, 2010 Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.

## Probability Biased and Unbiased – AIME 2010

Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability $\frac{4}{7}$,Ramesh flips the three coins, and then Suresh flips the three coins, let $\frac{m}{n}$ be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME, 2010, Question 4

Combinatorics by Brualdi

## Try with Hints

No heads TTT is $\frac{1.1.1}{2.2.7}=\frac{3}{28}$and $(\frac{3}{28})^{2}=\frac{9}{784}$

One Head HTT THT TTH with $\frac{3}{28}$ $\frac {3}{28}$ and $\frac{4}{28}$ then probability is $\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}$=$\frac{100}{784}$

Two heads HHT $\frac{4}{28}$ HTH $\frac{4}{28}$ THH $\frac{3}{28}$ then probability is $\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}$=$\frac{121}{784}$.

Three heads HHH is $\frac{4}{28}$ then probability $\frac{16}{784}$

Then sum is $\frac{9+100+121+16}{784}=\frac{123}{392}$ then 123+392=515.

Categories

## Probability | AMC 8, 2010 | Problem no. 24

Try this beautiful problem from Probability .You may use sequential hints to solve the problem.

## Probability | AMC-8, 2007 |Problem 24

A bag contains four pieces of paper, each labeled with one of the digits 1,2,3 or 4.  with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of 3?

• $\frac{3}{4}$
• $\frac{1}{2}$
• $\frac{1}{4}$

### Key Concepts

probability

combination

Number counting

Answer: $\frac{1}{2}$

AMC-8, 2007 problem 24

Challenges and Thrills in Pre College Mathematics

## Try with Hints

there are two ways that The combination of digits that give multiples of 3

Can you now finish the problem ……….

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4)

can you finish the problem……..

The combination of digits that give multiples of 3 are (1,2,3) and (2,3,4) . The number of ways to choose three digits out of four is 4. Therefore, the probability is $\frac{1}{2}$