You are more likely to encounter a dog in at least 1 out of _____ houses you visit.

• A. 4
• B. 3
• C. 2
• D. 5

Answer: (B)

When will you be more likely than not to run into a dog if you visit several houses? The idea is to compare the chance of not seeing a dog at all with the chance of seeing at least one dog.

Assume about 25% of households have a dog, so the chance a single house does not have a dog is 0.75. If you visit N independent houses, the probability that none of them has a dog is 0.75^N. You want the probability of at least one dog (the complement) to be greater than 50%, so 1 − 0.75^N > 0.5, which means 0.75^N < 0.5.

Solving with logarithms gives N > log(0.5)/log(0.75) ≈ 2.41. The smallest whole number that satisfies this is 3, so you’re more likely than not to encounter a dog in at least one of three houses you visit. If you visited two houses, the chance would be 1 − 0.75^2 = 0.4375, which is below 0.5, so three is the threshold. Visiting four houses would also exceed 0.5, but three is the minimum number needed to cross the threshold given this prevalence.