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AP Statistics 4.3 Introduction to Probability - MCQs - Exam Style Questions

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Question

Franklin will conduct a simulation by using a fair twelve-sided die that is used for a certain board game. For each roll of the die, there are 12 possible outcomes—the integers \(1\) through \(12\). For each of the first three trials of the simulation, Franklin rolled the twelve-sided die and each outcome of the roll was the number \(7\).
Based on the first three trials of the simulation, which of the following statements is true?
(A) On Franklin’s next roll of the die, the probability that the outcome of the roll will be the number \(7\) is greater than that of any of the other \(11\) possible outcomes.
(B) On Franklin’s next roll of the die, the probability that the outcome of the roll will be the number \(7\) is less than that of any of the other \(11\) possible outcomes.
(C) On Franklin’s next roll of the die, the probability that the outcome of the roll will be the number \(7\) is \(\tfrac13\).
(D) On Franklin’s next roll of the die, the probability that the outcome of the roll will be the number \(7\) is \(\tfrac{1}{12}\).
(E) On Franklin’s next roll of the die, the probability that the outcome of the roll will be the number \(7\) is \(\tfrac14\).
▶️ Answer/Explanation
Detailed solution
The die is fair and rolls are independent.
Past outcomes (three 7s in a row) do not affect the probability of the next outcome (no “streak effect”).
Each face has probability \( \tfrac{1}{12} \) on any single roll.
Therefore the probability that the next roll is \(7\) is \( \tfrac{1}{12} \).
✅ Answer: (D)

Question

According to a report for veterinarians in the United States, 36.5 percent of households in the United States own dogs and 30.4 percent of households in the United States own cats. If one household in the United States is selected at random, what is the probability that the selected household will own a dog or a cat?
(A) 0.111
(B) 0.331
(C) 0.558
(D) 0.669
(E) Not enough information is given to determine the probability.
▶️ Answer/Explanation
Detailed solution

1. General Addition Rule:
The probability of event A or B is:
$P(A \text{ or } B) = P(A) + P(B) – P(A \text{ and } B)$

2. Applying the Rule:
Let $P(\text{Dog}) = 0.365$ and $P(\text{Cat}) = 0.304$.
$P(\text{Dog or Cat}) = P(\text{Dog}) + P(\text{Cat}) – P(\text{Dog and Cat})$

3. Analysis:
The problem does not state that owning a dog and owning a cat are mutually exclusive events. A household could own both. Therefore, we do not know the value of $P(\text{Dog and Cat})$.

Without this information, the probability cannot be calculated.
✅ Answer: (E)

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