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Probability of an event- AMC 8 2009 Problem 13

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What are we learning ?

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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.

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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/Amc8_1102.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="4.3.1" 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 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

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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"]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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