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AMC 10 Australian Math Competition USA Math Olympiad

Statistics Problem – Australian Mathematics Competition, 2014

Problem No – 21 – Australian Mathematics Competition – 2014


Here is a problem based on Statistics from Australian Mathematics Competition, 2014.

In a competition between four people, Sally scored twice as many points as Brian and 30 points more than Corrie. Donna scored 50 points more than Brian. Which of the following statements is definitely true?

  • Sally won the competition.
  • Brian came last in the competition.
  • Donna won the competition.
  • Corrie beat Brian.
  • Sally and Donna together scored more than Brian and Corrie.

Key Concepts


Mathematical Analysis

Statistics

Arithmetic

Check the Answer


Answer: Sally and Donna together scored more than Brian and Corrie.

Australian Mathematics Competition – Upper Primary Division – 2014 – UP 21

Statistics 10th Edition – Robert S. Witte and John S. Witte

Try with Hints


For first hint we can use a table form with some possibilities :

statistics problem - solution

So in the previous hint from the table its clear that the 1st , 2nd and 4th options are not correct. Now if we go with the higher numbers like 90,100 etc then there will be a change with the values.Try to find it out using a table.

I guess you have noticed the differences in values let try to do that……

statistics problem - solution

So can understand from this table that the 3rd option is also not correct .

We are told that Sally scored 30 points more than Corrie and Donna scored 50 points more than Brian, and so together Sally and Donna always scored 80 points more than Corrie and Brian

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AMC 8 USA Math Olympiad

Probability AMC 10B 2019 problem 17

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What are we learning ?

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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.

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First look at the knowledge graph:-

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/02/p17.png” alt=”calculation of mean and median- AMC 8 2013 Problem” title_text=” mean and median- AMC 8 2013 Problem” align=”center” force_fullwidth=”on” _builder_version=”4.2.2″ min_height=”429px” height=”189px” max_height=”198px” custom_padding=”10px|10px|10px|10px|false|false”][/et_pb_image][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″ inline_fonts=”Aclonica”]

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 

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Start with hints 

[/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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