Question

1. your friend writes a phone number down on a piece of paper but the last three numt get smudged after being in your pocket all day long. you decide to randomly choose numbers for each of the three digits. use a sample space to determine whether guess the first digit correctly and the second digit correctly are independent events.

Explanation:

Step1: Define Events and Sample Space

Let the correct first digit be \( c_1 \) and correct second digit be \( c_2 \). Each digit (0 - 9) is equally likely. The sample space for one digit is \( S=\{0,1,2,\dots,9\} \), so for two digits, the sample space \( S_{2}=\{(d_1,d_2):d_1,d_2\in\{0,1,\dots,9\}\} \), with \( 10\times10 = 100 \) outcomes. Let event \( A \): first digit correct (\( d_1 = c_1 \)), event \( B \): second digit correct (\( d_2 = c_2 \)).

Step2: Calculate Probabilities

• \( P(A) \): Number of favorable for \( A \): 10 (since \( d_2 \) can be anything, \( d_1 = c_1 \)), so \( P(A)=\frac{10}{100}=0.1 \).
• \( P(B) \): Similarly, \( P(B)=\frac{10}{100}=0.1 \).
• \( P(A\cap B) \): Only 1 outcome (\( d_1 = c_1,d_2 = c_2 \)), so \( P(A\cap B)=\frac{1}{100}=0.01 \).

Step3: Check Independence

For independence, \( P(A\cap B)=P(A)\times P(B) \). \( P(A)\times P(B)=0.1\times0.1 = 0.01 \), which equals \( P(A\cap B) \).

Answer:

The events of guessing the first digit correctly and the second digit correctly are independent.