The Perils of Dogsitting: A Probability Warm-Up

Welcome to Real Numbers, a weekly show about discovering math by analyzing and solving real-world problems. I’m Sawyer. Each episode presents a problem of the week; in the following episode we discuss different ways to solve it.

Your friend is away for five nights and asks you to look after his poodle, Fluffy. Once your friend is far away, you learn that Fluffy wakes you up and demands a walk on two-thirds of nights. Assume the nights are independent: what is the probability that you get at least two full nights of sleep during the five nights? Show your work when you submit an answer.

The rest of this episode reviews probability, the mathematics of random or uncertain events. It connects to statistics, combinatorics and discrete mathematics, and gives us a way to model questions about an unknown future.

Consider a fair coin and a fair die. Heads and tails each have probability 1/2; each of the six numbers on the die has probability 1/6. If we flip the coin and roll the die together, what is the chance of heads and a 1?

The two events are independent: learning the result of one does not change the probabilities of the other. For example, revealing that the coin landed tails does not change the die’s distribution. We therefore multiply the probabilities of heads and 1: 1/2 × 1/6 = 1/12.

Now ask whether the next die roll is a 5 or a 6. Those events are disjoint: one roll cannot be both numbers. Their probabilities add, giving 1/6 + 1/6 = 1/3.

Heads or a 1 is different because the events can both happen. For any events A and B, P(A or B) = P(A) + P(B) − P(A and B). A Venn diagram shows why: adding both circles counts their overlapping region twice, so one copy must be subtracted.

For the coin and die, the result is 1/2 + 1/6 − 1/12 = 7/12. There is another way to calculate it using complements: for any event A, P(not A) = 1 − P(A), because exactly one of A and not A occurs.

The complement of heads or a 1 is tails and a number other than 1. Independence gives P(heads or 1) = 1 − (1/2 × 5/6) = 7/12, agreeing with the first method.

Use these rules carefully on the problem of the week: if Fluffy independently has a 2/3 chance of waking you each night, what is the chance that at least two of the five nights are uninterrupted? Good luck, and remember to show your work. We will discuss a solution next week.