2009 AMC 10B Problem 19

The clock is correct exactly when no digit $1$ appears anywhere in the display. Check the hour and the minutes separately.

1. Hours: the clock shows $1, 2, 3, \ldots, 12$. The ones with a $1$ in them are $1, 10, 11, 12$. That leaves $8$ good hours out of $12$, a fraction of $\frac{8}{12} = \frac23$.
2. Minutes: they go from $00$ to $59$.
- The tens digit can be $0, 1, 2, 3, 4, 5$. All but $1$ are good: $5$ choices.
- The ones digit can be $0$ to $9$. All but $1$ are good: $9$ choices.
- So $5 \times 9 = 45$ good minutes out of $60$, a fraction of $\frac{45}{60} = \frac34$.
3. Every hour has all $60$ minutes. So during a good hour, $\frac34$ of the time is shown correctly, and during a bad hour none of it is.
4. The fraction of the day with the correct time is
$$\tfrac23 \times \tfrac34 = \tfrac{6}{12} = \tfrac12.$$

The answer is $\boxed{\textbf{(A)}\ \frac12}$.

When two separate conditions must both hold, and every choice for one goes with every choice for the other, the fraction of good cases is the product of the two fractions. For the minutes you can count digit by digit, but the hours are not a simple digit pattern, so just list the bad hours.

Count the fraction of the day that is wrong. Bad hours: $4$ of $12$, which is $\frac13$ of the day. In the other $\frac23$ of the day, a quarter of the minutes are bad ($15$ of $60$), which is another $\frac23 \times \frac14 = \frac16$ of the day. Wrong time: $\frac13 + \frac16 = \frac12$, so right time: $\frac12$.