Difference between revisions of "1985 AHSME Problems/Problem 24"
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==Problem== | ==Problem== | ||
− | 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> \frac{1}{2} </math> the probability that the digit is | + | 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> \frac{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> | ||
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==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}} |
Revision as of 01:09, 3 April 2018
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
Notice that . Therefore, the probability of choosing is . The probability that the digit is chosen out of the set is twice that, .
which is the probability that the digit is from the set .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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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