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Use what you have learned about mutually inclusive and exclusive events.
Apply the formula for calculating the probability of events A or B.
There are four presidential candidates in a country. A street survey was conducted asking 2,500 registered voters which candidate they will vote for in the coming election. The table summarizes the results of the survey.
Presidential Candidates Number of Supporters
Candidate 1 250
Candidate 2 1,250
Candidate 3 450
Candidate 4 550
Part 1: What is the probability of randomly selecting a person who will vote for Candidate 1 or 2?
(2 points)
Responses
2502500
250 over 2500
15002500
1500 over 2500
10002500
1000 over 2500
12502500
1250 over 2500
Question 2
Part 2: If 1,000 more voters are surveyed, how many of them will likely prefer Candidate 3 or 4? Show your work and explain your answer.(2 points)
Apply the formula for calculating the probability of events A or B.
There are four presidential candidates in a country. A street survey was conducted asking 2,500 registered voters which candidate they will vote for in the coming election. The table summarizes the results of the survey.
Presidential Candidates Number of Supporters
Candidate 1 250
Candidate 2 1,250
Candidate 3 450
Candidate 4 550
Part 1: What is the probability of randomly selecting a person who will vote for Candidate 1 or 2?
(2 points)
Responses
2502500
250 over 2500
15002500
1500 over 2500
10002500
1000 over 2500
12502500
1250 over 2500
Question 2
Part 2: If 1,000 more voters are surveyed, how many of them will likely prefer Candidate 3 or 4? Show your work and explain your answer.(2 points)
Part 1:
To calculate the probability of randomly selecting a person who will vote for Candidate 1 or Candidate 2, we need to add the number of supporters for Candidate 1 and Candidate 2 and divide it by the total number of voters surveyed.
Number of supporters for Candidate 1 = 250
Number of supporters for Candidate 2 = 1,250
Total number of voters surveyed = 2,500
P(A or B) = P(A) + P(B) - P(A and B)
P(Candidate 1 or Candidate 2) = (250 + 1250) / 2500 = 1500 / 2500 = 0.6 or 60%
Therefore, the probability of randomly selecting a person who will vote for Candidate 1 or Candidate 2 is 60%.
Part 2:
If 1,000 more voters are surveyed, we can calculate how many of them will likely prefer Candidate 3 or Candidate 4 by using the same formula as above but incorporating the new total number of voters surveyed.
Number of supporters for Candidate 3 = 450
Number of supporters for Candidate 4 = 550
Total number of voters surveyed = 2,500 + 1,000 = 3,500
P(Candidate 3 or Candidate 4) = (450 + 550) / 3500 = 1000 / 3500 = 0.2857 or 28.57%
Therefore, if 1,000 more voters are surveyed, approximately 285.7 of them will likely prefer Candidate 3 or Candidate 4.
To calculate the probability of randomly selecting a person who will vote for Candidate 1 or Candidate 2, we need to add the number of supporters for Candidate 1 and Candidate 2 and divide it by the total number of voters surveyed.
Number of supporters for Candidate 1 = 250
Number of supporters for Candidate 2 = 1,250
Total number of voters surveyed = 2,500
P(A or B) = P(A) + P(B) - P(A and B)
P(Candidate 1 or Candidate 2) = (250 + 1250) / 2500 = 1500 / 2500 = 0.6 or 60%
Therefore, the probability of randomly selecting a person who will vote for Candidate 1 or Candidate 2 is 60%.
Part 2:
If 1,000 more voters are surveyed, we can calculate how many of them will likely prefer Candidate 3 or Candidate 4 by using the same formula as above but incorporating the new total number of voters surveyed.
Number of supporters for Candidate 3 = 450
Number of supporters for Candidate 4 = 550
Total number of voters surveyed = 2,500 + 1,000 = 3,500
P(Candidate 3 or Candidate 4) = (450 + 550) / 3500 = 1000 / 3500 = 0.2857 or 28.57%
Therefore, if 1,000 more voters are surveyed, approximately 285.7 of them will likely prefer Candidate 3 or Candidate 4.
Part 1: A basket of fruits contains 20 apples, 4 watermelon slices, and 12 oranges. If a piece of fruit is to be selected randomly from the box, what is the probability of selecting an apple or an orange?(2 points)
Responses
3236
32 over 36
2036
20 over 36
1236
12 over 36
1636
16 over 36
Question 2
Part 2: Explain the process you used to solve this problem. Show your work and explain using math vocabulary. (2 points)
Responses
3236
32 over 36
2036
20 over 36
1236
12 over 36
1636
16 over 36
Question 2
Part 2: Explain the process you used to solve this problem. Show your work and explain using math vocabulary. (2 points)
Part 1:
To calculate the probability of selecting an apple or an orange from the basket of fruits, we need to add the number of apples and the number of oranges and divide it by the total number of fruits in the basket.
Number of apples = 20
Number of oranges = 12
Total number of fruits = 20 (apples) + 4 (watermelon slices) + 12 (oranges) = 36
P(apple or orange) = (20 + 12) / 36 = 32 / 36 = 8 / 9 = 0.8889 or 88.89%
Therefore, the probability of randomly selecting an apple or an orange from the basket of fruits is approximately 88.89%.
Part 2:
To solve this problem, I first identified the total number of apples and oranges in the basket and then found the total number of fruits in the basket. I then added the number of apples and oranges together to find the total favorable outcomes. Finally, I divided the total favorable outcomes by the total number of fruits in the basket to find the probability of selecting an apple or an orange.
I used the concept of mutually inclusive events, where selecting an apple and selecting an orange are two separate events that can both result in a favorable outcome for the question of selecting either an apple or an orange. By adding the number of apples and oranges together, I ensured that I covered all possible favorable outcomes for the desired event.
To calculate the probability of selecting an apple or an orange from the basket of fruits, we need to add the number of apples and the number of oranges and divide it by the total number of fruits in the basket.
Number of apples = 20
Number of oranges = 12
Total number of fruits = 20 (apples) + 4 (watermelon slices) + 12 (oranges) = 36
P(apple or orange) = (20 + 12) / 36 = 32 / 36 = 8 / 9 = 0.8889 or 88.89%
Therefore, the probability of randomly selecting an apple or an orange from the basket of fruits is approximately 88.89%.
Part 2:
To solve this problem, I first identified the total number of apples and oranges in the basket and then found the total number of fruits in the basket. I then added the number of apples and oranges together to find the total favorable outcomes. Finally, I divided the total favorable outcomes by the total number of fruits in the basket to find the probability of selecting an apple or an orange.
I used the concept of mutually inclusive events, where selecting an apple and selecting an orange are two separate events that can both result in a favorable outcome for the question of selecting either an apple or an orange. By adding the number of apples and oranges together, I ensured that I covered all possible favorable outcomes for the desired event.
Part 1: A survey of 80 employees was conducted asking about which meals they regularly eat at the office. 20 employees eat breakfast at the office, 50 eat lunch, and 10 eat both breakfast and lunch. If an employee is randomly selected from this group, what is the probability of randomly selecting an employee who eat breakfast or lunch at the office?(2 points)
Responses
98
9 eighths
58
5 eighths
78
7 eighths
28
2 eighths
Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. (2 points)
Responses
98
9 eighths
58
5 eighths
78
7 eighths
28
2 eighths
Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. (2 points)
Part 1:
To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, we can use the formula:
P(Breakfast or Lunch) = P(Breakfast) + P(Lunch) - P(Breakfast and Lunch)
Given:
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
P(Breakfast or Lunch) = (20 + 50 - 10) / 80 = 60 / 80 = 0.75 or 75%
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 75%.
Part 2:
To arrive at this answer, I first identified the total number of employees who eat breakfast and lunch individually. Then, I subtracted the number of employees who eat both breakfast and lunch so as not to double-count them in the total favorable outcomes. Finally, I divided the sum of the number of employees who eat breakfast and lunch and subtracted the overlap by the total number of employees surveyed to get the probability of selecting an employee who eats breakfast or lunch.
I utilized the concept of mutually exclusive events here, as an employee can only eat one meal at a time. By subtracting the overlap of employees who eat both breakfast and lunch, I ensured that they were not counted twice in the calculation.
To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, we can use the formula:
P(Breakfast or Lunch) = P(Breakfast) + P(Lunch) - P(Breakfast and Lunch)
Given:
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
P(Breakfast or Lunch) = (20 + 50 - 10) / 80 = 60 / 80 = 0.75 or 75%
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 75%.
Part 2:
To arrive at this answer, I first identified the total number of employees who eat breakfast and lunch individually. Then, I subtracted the number of employees who eat both breakfast and lunch so as not to double-count them in the total favorable outcomes. Finally, I divided the sum of the number of employees who eat breakfast and lunch and subtracted the overlap by the total number of employees surveyed to get the probability of selecting an employee who eats breakfast or lunch.
I utilized the concept of mutually exclusive events here, as an employee can only eat one meal at a time. By subtracting the overlap of employees who eat both breakfast and lunch, I ensured that they were not counted twice in the calculation.