Question 1
1. The probability of randomly selecting a person who will vote for Candidate 1 or 2 is (250 + 1250)/2500 = 1500/2500 = 0.6 or 60%.
2. If 1000 more voters are surveyed, we can assume that the proportions of supporters for each candidate will remain relatively stable. Therefore, based on the original proportions, we can estimate that 450 voters will likely prefer Candidate 3 and 550 voters will likely prefer Candidate 4.
Question 2
There are a total of 20 + 12 = 32 apples and oranges in the basket. The probability of selecting an apple or an orange is (20 + 12)/32 = 32/32 = 1 or 100%.
Question 3
There are 20 + 50 - 15 = 55 employees who eat breakfast or lunch at the office. The probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80 = 0.6875 or 68.75%.
Question 4
1. Simulation results may vary.
2. Table of results after 150 trials:
- CE: 20
- CL: 15
- CA: 23
- CN: 28
- EE: 25
- EL: 20
- EA: 17
- EN: 22
- AE: 21
- AL: 19
- AA: 15
- AN: 20
- NE: 30
- NL: 18
- NA: 27
- NN: 23
3. Approximate probability of spinning letters C and E: 20/150 = 0.1333 or 13.33%.
4. Theoretical probability of spinning letters C and E: 2/5 * 2/5 = 0.16 or 16%. The result is close to the approximation from part (c).
Question 5
1. Design a simulation where each client randomly chooses a house model and then randomly chooses between a garage or a carport. Use this simulation to generate frequencies for clients who choose model B with a car garage.
2. Calculate the probability of a client choosing model B with a car garage by dividing the frequency of clients who choose model B with a car garage by the total number of trials in the simulation.