AP Statistics 4.6 Independent Events and Unions of Events- MCQs - Exam Style Questions
Question
Suppose study participants are asked two questions. Based on a large number of participants, the probability distribution for the responses to Questions 1 and 2 is given below.
Question 1 | |||
---|---|---|---|
Correct | Incorrect | ||
Question 2 | Correct | 0.25 | 0.20 |
Incorrect | 0.08 | 0.47 |
For one participant, let $A=$ Question 1 is answered correctly and $B=$ Question 2 is answered correctly. Which of the following must be true?
(B) A and B are not independent because the probability in each cell is not 0.25.
(C) A and B are not independent because $(0.33)(0.45)\ne0.25$
(D) A and B are independent because one participant’s answers do not affect another participant’s answers.
(E) There is not enough information provided to determine whether A and B are independent.
▶️ Answer/Explanation
Two events, A and B, are independent if the probability that they both occur is equal to the product of their individual probabilities. The rule is: $P(A \text{ and } B) = P(A) \times P(B)$.
Let’s find the probabilities from the table:
– $P(A)$ = Probability that Question 1 is correct. We add the probabilities in the “Correct” column for Question 1: $0.25 + 0.08 = 0.33$.
– $P(B)$ = Probability that Question 2 is correct. We add the probabilities in the “Correct” row for Question 2: $0.25 + 0.20 = 0.45$.
– $P(A \text{ and } B)$ = Probability that both are correct. This is given directly in the table: $0.25$.
Now, let’s check if the independence rule holds:
Is $P(A) \times P(B) = P(A \text{ and } B)$?
$0.33 \times 0.45 = 0.1485$
Since $0.1485$ is not equal to $0.25$, the events are not independent. Option (C) correctly states this reasoning.
✅ Answer: (C)
Question
(B) \(0.45\)
(C) \(0.50\)
(D) \(0.55\)
(E) \(0.60\)
▶️ Answer/Explanation
1. Define Events:
Let \(R_1\) = first light red, \(R_2\) = second light red
Given: \(P(R_1 \cap R_2) = 0.22\), \(P(R_1 \cap R_2^c) = 0.33\)
2. Find \(P(R_1)\):
\(P(R_1) = P(R_1 \cap R_2) + P(R_1 \cap R_2^c) = 0.22 + 0.33 = 0.55\)
3. Use Independence:
Since lights are independent: \(P(R_1 \cap R_2) = P(R_1) \times P(R_2)\)
\(0.22 = 0.55 \times P(R_2)\)
4. Solve for \(P(R_2)\):
\(P(R_2) = \frac{0.22}{0.55} = 0.40\)
This matches option (A).
✅ Answer: (A)