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Problem 632

AMC 10/12, early questions

Algebra Difficulty 3.8 Multiple choice CEMC Cayley · Canada · 2015

Nylah has her living room lights on a timer. Each evening, the timer switches the lights on randomly at exactly 7:00 p.m., 7:30 p.m., 8:00 p.m., 8:30 p.m., or 9:00 p.m. Later in the evening, the timer switches the lights off at any random time between 11 p.m. and 1 a.m. For example, the lights could be switched on at exactly 7:30 p.m. and off at any one of the infinite number of possible times between 11 p.m. and 1 a.m. On a given night, Nylah’s lights are on for $t$ hours. What is the probability that $4 < t < 5$ ?

Pick one

Official solution

Nylah’s lights come on randomly at one of the times 7:00 p.m., 7:30 p.m., 8:00 p.m., 8:30 p.m., or 9:00 p.m., each with probability $\frac{1}{5}$ .

What is the probability that the lights come on at 7:00 p.m. and are on for $t$ hours with $4<t<5$ ?

If the lights come on at 7:00 p.m. and are on for between 4 and 5 hours, then they go off between 11:00 p.m. and 12:00 a.m.

Since the length of this interval is 1 hour and the length of the total interval of time in which the lights randomly go off is 2 hours (11:00 p.m. to 1:00 a.m.), then the probability that they go off between 11:00 p.m. and 12:00 a.m. is $\frac{1}{2}$ .

Therefore, the probability that the lights come on at 7:00 p.m. and are on for $t$ hours with $4<t<5$ is $\frac{1}{5}\cdot\frac{1}{2}=\frac{1}{10}$ .

Similarly, if the lights come on at 7:30 p.m., they can go off between 11:30 p.m. and 12:30 a.m., and the probability of this combination of events is also $\frac{1}{5}\cdot\frac{1}{2}=\frac{1}{10}$ .

Similarly again, the probability of the lights coming on at 8:00 p.m. and going off between 12:00 a.m. and 1:00 a.m. is also $\frac{1}{10}$ .

If the lights come on at 8:30 p.m., then to be on for between 4 and 5 hours, they must go off between 12:30 a.m. and 1:00 a.m. (They cannot stay on past 1:00 a.m.)

The probability of this combination is $\frac{1}{5}\cdot\frac{1/2}{2}=\frac{1}{5}\cdot\frac{1}{4}=\frac{1}{20}$ .

If the lights come on at 9:00 p.m., they cannot be on for more than 4 hours, since the latest that they can go off is 1:00 a.m.

Therefore, the probability that the lights are on for between 4 and 5 hours is $3\cdot \frac{1}{10}+\frac{1}{20}=\frac{7}{20}$ .

(We note that we can safely ignore the question of whether the lights coming on at 7:30 p.m. and going off at exactly 11:30 p.m., for example, affects the probability calculation, because 11:30 p.m. is a single point in an interval containing an infinite number of points, and so does not affect the probability.)

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.