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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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AMC 8 Combinatorics Math Olympiad Probability

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?

Problem from Probability

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

Key Concepts


probability

Equilly likely

Number counting

Check the Answer


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…….

\(1 – \frac{2}{4} \times \frac{2}{3}\)= \(1 – \frac{1}{3} = \frac{2}{3} \)  

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AMC 8 Combinatorics Math Olympiad Probability USA Math Olympiad

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

Check the Answer


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 \(\frac{6}{16}=\frac{3}{8}\)

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AMC 8 Combinatorics Math Olympiad Probability

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

Check the Answer


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

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AMC 8 USA Math Olympiad

Combinatorics : AMC 8 2008 Problem 14

What is Combinatorics?


Combinatorics is a field of Mathematics where we study in how many ways we can arrange some number of given objects following some certain rules of arrangements.

Try the problem


Three $\text{A’s}$, three $\text{B’s}$, and three $\text{C’s}$ are placed in the nine spaces so that each row and column contain one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

AMC 8 , 2008 Problem 14

Combinatoric

6 out of 10

Mathematical Circles

Knowledge Graph


Use some hints


Lets solve the problem together !!!

lets name all the small boxes : We will call the box on the intersection of $i^{th}$ row and $j^{th}$ column $\text{Box}(i,j)$

$1$ of the three A’s is fixed in the $\text{Box}(1,1)$

Then A can not occur in the $1^{st}$ row and $1^{st}$ column.

So to place a B in the $1^{st}$ row we have 2 choices[ $\text{Box}(1,2)$ and $\text{Box}(1,3)$ ] and then only $1$ choice left to place C in the $1^{st}$ row.

Now can you think about the placements of A, B and C is the second row ??

A can not occur in the $1^{st}$ column so there are $2$ choices to place A in $2^{nd}$ row and then only one choice left to place C in $2^{nd}$ row.

Now think about the $3^{rd}$ row.

After placing all the letters in the described manner we are automatically left with only one way to place A, B and C in the third row.

So the total number of choice in $1^{st}$ row is $2$ for each of these ways there are $2$ choices for the $2^{nd}$ row and for each of these $2\times 2$ choices there is only one choice for the $3^{rd}$ row.

Hence the total number of choices : $2 \times 2 \times 1=4$.

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AMC 10 USA Math Olympiad

Permutation – AMC 10B – 2020 – Problem No.5

What is Permutation ?


Permutation is the act of arranging the members of a set into a sequence or order, or, if the set is already ordered, rearranging (reordering) its elements—a process called permuting.

Try the problem from AMC 10B – 2020 – Problem 5


How many distinguishable arrangements are there of  1 brown tile,1 purple tile ,2 green tiles and 3 yellow tiles in a row from left to right ? (Tiles of the same color are indistinguishable.)

A) 210 B) 420 C) 630 D) 840 E) 1050

American Mathematics Competition 10 (AMC 10B), 2020, Problem Number – 5

Permutation

5 out of 10

Mathematical Circle

Knowledge Graph


Permutation- knowledge graph

Use some hints


If you really need a hint you can go through the concept of probability at first : In its simplest form, probability can be expressed mathematically as: the number of occurrences of a targeted event divided by the number of occurrences plus the number of failures of occurrences (this adds up to the total of possible outcomes):

\(p(a) = \frac {p(a)} { [P(a) + p(b)] } \)

Let’s try to find how many possibilities there would be if they were all distinguishable, then divide out the ones we over counted . There are  7! ways to order 7 objects. However, since there’s 3!= 6 ways to switch the yellow tiles around without changing anything (since they’re indistinguishable) and  2! = 2 ways for green tiles.

I am sure that you are almost there for the final calculation but let me help those who are still not there

\(\frac {7!}{6 \cdot 2} = 420 \) .

So the correct answer is B) 420

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AMC 10 USA Math Olympiad

Probability Problem from AMC 10A – 2020 – Problem No. 15

What is Probability?


The Probability theory, a branch of mathematics concerned with the analysis of random phenomena. The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. The actual outcome is considered to be determined by chance.

Try the Problem from AMC 10 – 2020


A positive integer divisor of 12! is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as \(\frac {m}{n}\), where m and n are relatively prime positive integers. What is m+n ?

A)3 B) 5 C)12 D) 18 E) 23

American Mathematics Competition 10 (AMC 10), {2020}, {Problem number 15}

Inequality (AM-GM)

6 out of 10

Secrets in Inequalities.

Knowledge Graph


Probability- knowledge graph

Use some hints


If you really need any hint try this out:

The prime factorization of  12! is \(2^{10}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11\)

This yields a total of  \( 11\cdot 6 \cdot 3 \cdot 2 \cdot 2 \) divisors of 12!.

In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization.

Again 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in 12!. Thus, there are \(6 \cdot 3\cdot 2\) perfect squares.

I think you already got the answer but if you have any doubt use the last hint :

So the probability that the divisor chosen is a perfect square is \(\frac {6.3 . 2}{11 . 6. 3. 2. 2} = \frac {1}{22}\)

\(\frac {m}{n} = \frac {1}{22} \)

m+n = 1+22 = 23.

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AMC 10 USA Math Olympiad

Combinatorics – AMC 10A 2008 Problem 23 Sequential Hints

[et_pb_section fb_built=”1″ _builder_version=”4.0″][et_pb_row _builder_version=”3.25″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Understand the problem

[/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway||||||||” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Two subsets of the set $S=\lbrace a,b,c,d,e\rbrace$ are to be chosen so that their union is $S$ and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter? $\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.0″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.0″ open=”on”]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]

Combinatorics 

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” open=”off” _builder_version=”4.0″]

7/10

[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]

Enumetarive Combinatorics – ( Problem Solving Strategies ) by Arthur Engel  

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Start with hints

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Watch video

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Connected Program at Cheenta

[/et_pb_text][et_pb_tabs active_tab_background_color=”#0c71c3″ inactive_tab_background_color=”#000000″ _builder_version=”4.0″ tab_text_color=”#ffffff” tab_font=”||||||||” background_color=”#ffffff” custom_padding=”||153px|25px||”][et_pb_tab title=”Hint 0″ _builder_version=”3.22.4″]You could give it a thought first…are you sure you really need a hint ?

[/et_pb_tab][et_pb_tab title=”Hint 1″ _builder_version=”4.0″]

 First can you try finding the number of ways in which we can choose the 2 shared elements ?  That would be 5C2 = 10 ways. Can you try completing this ?    

[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0″]

 All that remains now is to place the remaining 3 elements into the subsets. In how many ways can we do this ?

[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0″]

So basically, if we think about it a little, it is equivalent to the problem of placing 3 elements in 4 gaps. Evidently, this can be done in 4C3 = 4 ways.  Now, it’s pretty easy to arrive at the answer…could you do it by yourself ?                        

[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0″]

The final answer can be easily found out using the Multiplication Principle, which leads us to… 4 x 10 = 40 ways 

And, we are done !

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/matholympiad/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”3.23.3″ custom_button=”on” button_bg_color=”#0c71c3″ button_border_color=”#0c71c3″ button_border_radius=”0px” button_font=”Raleway||||||||” button_icon=”%%3%%” background_layout=”dark” button_text_shadow_style=”preset1″ box_shadow_style=”preset1″ box_shadow_color=”#0c71c3″][/et_pb_button][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Similar Problems

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