Answer:
1/16
Step-by-step explanation:
We can look at this problem in the same way we'd solve probability questions. What's the likelihood that both of the hands of the clock point in the same quarter of the clock?
For either hand, it has a 25% (or 1/4) probability of being at this quarter of the clock.
Therefore, if we want the first hand AND the second hand to be in the same quarter of the clock, the probability of that would be [tex]\frac14 \times \frac14 = \left(\frac14\right)^2 = \frac1{16}[/tex]
Therefore, both hands of the clock are not visible for 1/16th of the day.
