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AMC-8 USA Math Olympiad

Probability of an event- AMC 8 2009 Problem 13

Try this beautiful problem from AMC 8. It involves probability and divisibility. We provide sequential hints so that you can try the problem.

What are we learning ?

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.

First look at the knowledge graph:-

calculation of  mean and median- AMC 8 2013 Problem

Next understand the problem

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 $5$? $\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$
Source of the problem

American Mathematical Contest 2009, AMC 8 Problem 13

Key Competency

Finding probability of an event using the concept of divisibility

Difficulty Level
5/10
Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

Start with hints 

Do you really need a hint? Try it first!
First note that a number is divisible by $5$ if the unit digit is $5$ or $0$
Can you find all the three digit numbers that can be made using the digits given  
So the three digit numbers are $135,153,351,315,513,531$. among these only two numbers are divisible by $5$ 
Probability of an event = $\frac{\textbf{Number points supporting the event}}{Total number of outcomes}$ Ans=$\frac{\textbf{Number of 3 digit numbers divisible by 5}}{\textbf{Number of 3 digits number can be made}}=\frac{2}{6}=\frac{1}{3}$

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