Difference between revisions of "1985 AHSME Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | A non-zero | + | A non-zero digit is chosen in such a way that the probability of choosing digit <math>d</math> is <math>\log_{10}{(d+1)}-\log_{10}{d}</math>. The probability that the digit <math>2</math> is chosen is exactly <math>1/2</math> the probability that the digit chosen is in the set |
− | <math> \mathrm{(A)\ } \{2, 3\} \qquad \mathrm{(B) \ }\{3, 4\} \qquad \mathrm{(C) \ } \{4, 5, 6, 7, 8\} \qquad \mathrm{(D) \ } \{5, 6, 7, 8, 9\} \qquad \mathrm{(E) \ }\{4, 5, 6, 7, 8, 9\} </math> | + | <math> \mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,6,7,8\} \qquad \mathrm{(D) \ } \{5,6,7,8,9\} \qquad \mathrm{(E) \ }\{4,5,6,7,8,9\} </math> |
==Solution== | ==Solution== | ||
− | + | We have <math>\log_{10}{(d+1)}-\log_{10}{d} = \log_{10}{\left(\frac{d+1}{d}\right)}</math>, so the probability of choosing <math>2</math> is <math>\log_{10}{\left(\frac{3}{2}\right)}</math>. The probability that the digit chosen is in the set must therefore be <cmath>\begin{align*}2\log_{10}{\left(\frac{3}{2}\right)} = &\log_{10}{\left(\left(\frac{3}{2}\right)^2\right)} \ = &\log_{10}{\left(\frac{9}{4}\right)} \ = &\log_{10}{9}-\log_{10}{4} \ = &\left(\log_{10}{9}-\log_{10}{8}\right)+\left(\log_{10}{8}-\log_{10}{7}\right)+\left(\log_{10}{7}-\log_{10}{6}\right) \ &+\left(\log_{10}{6}-\log_{10}{5}\right)+\left(\log_{10}{5}-\log_{10}{4}\right),\end{align*}</cmath> | |
− | + | which, by definition, is the probability that the digit chosen is in the set <math>\boxed{\text{(C)} \ \{4,5,6,7,8\}}</math>. | |
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− | which is the probability that the digit is | ||
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=23|num-a=25}} | {{AHSME box|year=1985|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:03, 20 March 2024
Problem
A non-zero digit is chosen in such a way that the probability of choosing digit is . The probability that the digit is chosen is exactly the probability that the digit chosen is in the set
Solution
We have , so the probability of choosing is . The probability that the digit chosen is in the set must therefore be which, by definition, is the probability that the digit chosen is in the set .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AHSME Problems and Solutions |
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