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Difference between revisions of "1985 AHSME Problems/Problem 6"
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==Problem==
 
==Problem==
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One student in a class of boys and girls is chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is <math> \frac{2}{3} </math> of the [[probability]] that a girl is chosen. The [[ratio]] of the number of boys to the total number of boys and girls is
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One student in a class of boys and girls is chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is <math>\frac{2}{3}</math> of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is
  
 
<math> \mathrm{(A)\ } \frac{1}{3} \qquad \mathrm{(B) \ }\frac{2}{5} \qquad \mathrm{(C) \  } \frac{1}{2} \qquad \mathrm{(D) \  } \frac{3}{5} \qquad \mathrm{(E) \  }\frac{2}{3}  </math>
 
<math> \mathrm{(A)\ } \frac{1}{3} \qquad \mathrm{(B) \ }\frac{2}{5} \qquad \mathrm{(C) \  } \frac{1}{2} \qquad \mathrm{(D) \  } \frac{3}{5} \qquad \mathrm{(E) \  }\frac{2}{3}  </math>
  
 
==Solution==
 
==Solution==
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Let the probability that a boy is chosen be <math> p </math>. Since the sum of the probability that a boy is chosen and a girl is chosen is <math> 1 </math>, we have the probability that a girl is chosen is <math> 1-p </math>. We know that <math> \frac{2}{3} </math> of this is <math> p </math>, so <math> \frac{2}{3}(1-p)=p\implies 2(1-p)=3p\implies 2-2p=3p\implies 5p=2\implies p=\frac{2}{5} </math>.
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Let the probability that a boy is chosen be <math>p</math>, so the probability that a girl is chosen is <math>1-p</math> (as the probabilities must sum to <math>1</math>). Therefore <cmath>\begin{align*}\frac{2}{3}(1-p) = p &\iff 2(1-p) = 3p \\ &\iff 2-2p = 3p \\&\iff 5p = 2 \\&\iff p = \frac{2}{5}.\end{align*}</cmath>
  
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Now simply notice that the probability a boy is chosen is precisely the proportion of the number of boys out of the total number of students (boys and girls), so the answer is precisely <math>\boxed{\text{(B)} \ \frac{2}{5}}</math>.
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Notice that <math> P(\text{A boy is chosen})=\frac{\text{Number of boys}}{\text{Number of boys}+\text{Number of girls}} </math>. But we already know that this is <math> \frac{2}{5} </math>, so that is our final answer, <math> \boxed{\text{B}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=5|num-a=7}}
 
{{AHSME box|year=1985|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:59, 19 March 2024

Problem

One student in a class of boys and girls is chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is $\frac{2}{3}$ of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is

$\mathrm{(A)\ } \frac{1}{3} \qquad \mathrm{(B) \ }\frac{2}{5} \qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ } \frac{3}{5} \qquad \mathrm{(E) \ }\frac{2}{3}$

Solution

Let the probability that a boy is chosen be $p$, so the probability that a girl is chosen is $1-p$ (as the probabilities must sum to $1$). Therefore \begin{align*}\frac{2}{3}(1-p) = p &\iff 2(1-p) = 3p \\ &\iff 2-2p = 3p \\&\iff 5p = 2 \\&\iff p = \frac{2}{5}.\end{align*}

Now simply notice that the probability a boy is chosen is precisely the proportion of the number of boys out of the total number of students (boys and girls), so the answer is precisely $\boxed{\text{(B)} \ \frac{2}{5}}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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