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## 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

## Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

## Complex Numbers – AIME, 2009

There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

• 101
• 201
• 301
• 697

### Key Concepts

Complex Numbers

Theory of equations

Polynomials

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

## Try with Hints

Taking z=a+bi

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

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## Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

## Combinations- AIME, 2009

A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to$9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

• 110
• 420
• 430
• 111

### Key Concepts

Combinations

Theory of equations

Polynomials

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

## Try with Hints

Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420