Difference between revisions of "1985 AHSME Problems/Problem 6"
Sevenoptimus (talk | contribs) m (Improved formatting) |
|||
Line 1: | Line 1: | ||
==Problem== | ==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 <math> \frac{2}{3} </math> of the | + | 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== | ||
− | Let the probability that a boy is chosen be <math> p </math> | + | 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> |
− | + | 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>. | |
− | |||
==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 17:59, 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 of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is
Solution
Let the probability that a boy is chosen be , so the probability that a girl is chosen is (as the probabilities must sum to ). Therefore
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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.