Question

1. a bag contains six pieces of paper, numbered 1 through 6. a student randomly selects a piece of paper, replaces it, and randomly selects another piece of paper. use a sample space to determine whether randomly selecting a 5 first and randomly selecting an odd number second are independent events.

Explanation:

Step1: Define Events

Let \( A \) be the event of selecting a 5 first, and \( B \) be the event of selecting an odd number second. The sample space \( S \) for two selections (with replacement) is \( S=\{(i,j):i,j\in\{1,2,3,4,5,6\}\} \), so \( n(S) = 6\times6=36 \).

Step2: Find \( P(A) \)

For event \( A \), the first number is 5, so \( A=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \), \( n(A) = 6 \). Thus, \( P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6} \).

Step3: Find \( P(B) \)

Odd numbers are 1,3,5. So \( B=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3),(5,5),(6,1),(6,3),(6,5)\} \), \( n(B) = 18 \). Thus, \( P(B)=\frac{n(B)}{n(S)}=\frac{18}{36}=\frac{1}{2} \).

Step4: Find \( P(A\cap B) \)

\( A\cap B=\{(5,1),(5,3),(5,5)\} \), \( n(A\cap B) = 3 \). Thus, \( P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{3}{36}=\frac{1}{12} \).

Step5: Check Independence

Two events are independent if \( P(A\cap B)=P(A)\times P(B) \). Calculate \( P(A)\times P(B)=\frac{1}{6}\times\frac{1}{2}=\frac{1}{12} \), which equals \( P(A\cap B) \).

Answer:

The events of randomly selecting a 5 first and randomly selecting an odd number second are independent events.