Difference between revisions of "2019 Mock AMC 10B Problems/Problem 6"

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==Solution==
 
==Solution==
  
Each die has <math>3</math> prime numbers: <math>2, 3, 5</math>. Since the numbers rolled on each die must be distinct, the answer is <math>\frac{3}{5} \cdot \frac{2}{6} = \boxed{\text{(C)} \frac{1}{5}}</math>.
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Each die has <math>3</math> prime numbers: <math>2, 3, 5</math>. Since the numbers rolled on each die must be distinct, the answer is <math>\frac{2}{5} \cdot \frac{3}{6} = \boxed{\text{(C)} \frac{1}{5}}</math>.
 
 
<baker77>
 

Latest revision as of 17:09, 27 August 2020

Problem

Mark rolled two standard dice. Given that he rolled two distinct values, find the probability that he rolled two primes.

$\textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{1}{7}\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{5}$

Solution

Each die has $3$ prime numbers: $2, 3, 5$. Since the numbers rolled on each die must be distinct, the answer is $\frac{2}{5} \cdot \frac{3}{6} = \boxed{\text{(C)} \frac{1}{5}}$.

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