1985 AMC 12 Problem 24

Question

A non-zero digit is chosen in such a way that the probability of choosing digit

d

is

\log_{10}(d+1)-\log_{10}d.

The probability that the digit

2

is chosen is exactly

\frac{1}{2}

the probability that the digit chosen is in the set

\{2,3\}

\{3,4\}

\{4,5,6,7,8\}

\{5,6,7,8,9\}

\{4,5,6,7,8,9\}

Twice the probability of digit

2

is

2\log_{10}(\frac{3}{2})

Sum the probabilities over a consecutive set by telescoping its logarithms

The probability of

2

is

\log_{10}(\frac{3}{2}),

so twice that probability is

\log_{10}(\frac{9}{4}).

For the digits

4

through

8,

the sum telescopes:

\begin{aligned} &\sum_{d=4}^{8} \bigl(\log_{10}(d+1)\\ &\qquad{}-\log_{10}d\bigr) =\log_{10}\frac94. \end{aligned}

Therefore that set has exactly twice the probability of digit

2.

Thus the correct answer is C.

Problem 24 in other years