Categories

## Competency in Focus: Probability of an event

This problem from American Mathematics Contest 8 (AMC 8, 2019) is based on calculation of probability of an event using the concept of divisibility. It is Question no. 13 of the AMC 8 2009 Problem series.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]A three-digit integer contains one of each of the digits $1$, $3$, and $5$. What is the probability that the integer is divisible by $5$? $\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$[/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=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.3.1″ hover_enabled=”0″]

American Mathematical Contest 2009, AMC 8 Problem 13 [/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.3.1″ hover_enabled=”0″ inline_fonts=”Abhaya Libre” open=”off”]

### Finding probability of an event using the concept of divisibility

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]5/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]First note that a number is divisible by $5$ if the unit digit is $5$ or $0$[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]Can you find all the three digit numbers that can be made using the digits given  [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]So the three digit numbers are $135,153,351,315,513,531$. among these only two numbers are divisible by $5$ [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Probability of an event = $\frac{\textbf{Number points supporting the event}}{Total number of outcomes}$ Ans=$\frac{\textbf{Number of 3 digit numbers divisible by 5}}{\textbf{Number of 3 digits number can be made}}=\frac{2}{6}=\frac{1}{3}$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]

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Categories

## 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.

## 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.

Categories

## Competency in Focus: Concept of Probability

This problem from American Mathematics Contest 8 (AMC 8, 2018) is based on calculation of probability. It is Question number 11 of the AMC 8 2018 Problem series.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″] Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown. If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column? $\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{2}{5} \qquad \textbf{(C) } \frac{7}{15} \qquad \textbf{(D) } \frac{1}{2} \qquad \textbf{(E) } \frac{2}{3}$  [/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=”4.3.1″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”off” _builder_version=”4.3.1″ hover_enabled=”0″]

### American Mathematical Contest 2018, AMC 8 Problem 11

[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.3.1″ hover_enabled=”0″ inline_fonts=”Abhaya Libre” open=”on”]

### Basic Probability sum

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]6/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]If you need any hint try from this: There are a total of $6!$ ways to arrange the kids.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]Abby and Bridget can sit in 3 ways if they are adjacent in the same column: For each of these seat positions, Abby and Bridget can switch seats, and the other 4 people can be arranged in $4!$ ways which results in a total of $3 \times 2 \times 4!$ ways to arrange them.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]By the same logic, there are 4 ways for Abby and Bridget to placed if they are adjacent in the same row, they can swap seats, and the other $4$ people can be arranged in $4!$ ways, for a total of $4 \times 2 \times 4!$ ways to arrange them.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]We sum the 2 possibilities up to get $\frac{(3\cdot2)\cdot4!+(4\cdot2)\cdot4!}{6!} = \frac{14\cdot4!}{6!}=\boxed{\frac{7}{15}}$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]

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Categories

## Competency in Focus: probability

This problem is from American Mathematics Contest 10B (AMC 10B, 2019). It is Question no. 17 of the AMC 10B 2019 Problem series.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3….$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$[/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=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”off”]American Mathematical Contest 2019, AMC 10B Problem 17[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]

### Probability

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”on”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + … = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]We can see that each time we add a bin between the two balls, the probability halves.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]

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Categories

# 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″]Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

[/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” open=”on” _builder_version=”4.0″]American Mathematics Competition [/et_pb_accordion_item][et_pb_accordion_item title=”Topic” _builder_version=”4.0″ open=”off”]Probability / Combinatorics

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0″ open=”off”]’7/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”3.29.2″ open=”off”]Challenges and Thrills of Pre-College Mathematics

[/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=”||150px|18px||”][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″]

Let’s not get worried by the idea of having to find out the probability. For, we may simply recall : Probability = Number of favorable events / Total number events in the sample space So, the denominator is easy to find out here, 8 people, all of whom are either standing or sitting. So, 2 possibilities for every person. And that makes our denominator, $2^8 = 256$ Now, with a little bit of focus, it is not at all hard to see that the numerator, “number of favorable events” is essentially a Combinatorics arrangement problem.

So, would you like to have a go at this by yourself now ?

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

Now, as we go ahead with the solution, let us develop the idea of the strategy we could use here. You must be familiar with recursion, but let’s just recall its philosophy. Recursion essentially meant that you could express a variable or a term in terms of its own parameters. As in, when a function calls itself. If you want a brief look-up, catch the short story in the paragraph below [ Much like imagine, you’re eating a large cake. You won’t really eat the whole thing at once, would you ? You would possibly enjoy it in parts. Imagine you’d have 1/3 rd of it, to begin with. Then if your problem initially is, EAT THE CAKE. Even, after you have had a first bite, as I just said, what’s the name of your problem ? It’s still EAT THE CAKE , isn’t it ? Yes, you’re still eating the same cake, only your problem is reduced by 1/3-rd, somewhat like a miniature version. Something we could call, ( EAT THE CAKE ) / 3…That’s pretty much, an intuitive idea of recursion. ] How could we model our problem in that fashion ? Imagine we talk of 8 persons standing in the line, to begin with. Let us try to develop a recurrence relation for the problem. Let’s talk of a person, N in the queue. See the catch here, if this person N is sitting, our problem size reduces by 1 – which means it is now equivalent to 7 people standing in the queue. On the other hand if this person N is standing, it is implied by the statement of the problem, that the adjacent people on his either side must be sitting ! That reduces our problem size by 2… Hey, so now that we have kind of linguistically framed it, could you try out a mathematical interpretation ?

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

Let, the answer we are looking for is a_n. ( That is, the number of ways to arrange n people in a line such that no two of them, who are adjacent, are standing ) So for a_n >= 2,
See there can be only two broad-spectrum cases : Case I : Person #1 in the line is standing If Person #1 in the line is standing, the second one is sitting. So the rest can be arranged in a_n-2 ways.  Case II : Person #1 in the line is sitting
This means the rest can be arranged in a_n-1 ways. So, Case I + Case II, adds up to the relation we are looking for ? That would be something like :  a_n = a_n-2 + a_n-1 Wait ! Does this, by any chance seem familiar ? Does it ? Well yes, this is nothing but the Fibonacci Recurrence Relation. The problem of rabbits in the field, if you need some off-beat reference !  [ The Fibonacci Sequence goes like :
1,1,2,3,5,8,13, 21, 34,….
It’s quite easy to see that from the third time onwards, every term is numerically equal to the sum of its two previous terms ] So, do you think you could conclude this…and find the required probability ?

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

So we just saw, we can map the given problem at hand to the Fibonacci problem.  Our starting cases here would be a_0 = 1 and a_1 = 2. ( By the problem ). This means our ans would be the (n+2)th Fibonacci number for a_n F_n stands for the n’th Fibonacci number, of the sequence you just came across in the last hint. So that gives us a_5 + a_7 = F_7 + F_9 = 13 + 34 = 47. So, that’s our numerator ! Hence, the required probability =   47 / 256

And that’s pretty much of all we need !