Question 1
a. What is the probability of randomly selecting a person who will vote for Candidate 1 or 2?
To find the probability of randomly selecting a person who will vote for Candidate 1 or Candidate 2, we first need to sum the number of supporters of both candidates.
- Number of supporters for Candidate 1: 250
- Number of supporters for Candidate 2: 1,250
Total supporters for Candidate 1 and 2 = 250 + 1,250 = 1,500
Next, we calculate the total number of registered voters surveyed, which is 2,500.
The probability \( P \) of selecting a person who will vote for Candidate 1 or 2 is given by the formula:
\[ P(\text{Candidate 1 or 2}) = \frac{\text{Number of supporters for Candidate 1 or 2}}{\text{Total number of voters}} = \frac{1,500}{2,500} \]
Calculating this gives:
\[ P(\text{Candidate 1 or 2}) = \frac{1,500}{2,500} = 0.6 \]
To express it as a percentage, we multiply by 100:
\[ 0.6 \times 100 = 60% \]
Answer: 60%
b. If 1,000 more voters are surveyed, how many of those 1,000 will likely prefer Candidate 3 or 4?
To predict how many of the additional 1,000 voters will prefer Candidate 3 or Candidate 4, we first find the total number of supporters for Candidates 3 and 4.
- Number of supporters for Candidate 3: 450
- Number of supporters for Candidate 4: 550
Total supporters for Candidate 3 and 4 = 450 + 550 = 1,000
Now, we need to calculate the probability that a randomly selected voter from the original survey will prefer Candidate 3 or 4:
\[ P(\text{Candidate 3 or 4}) = \frac{\text{Number of supporters for Candidate 3 or 4}}{\text{Total number of voters}} = \frac{1,000}{2,500} = 0.4 \]
Now, we use this probability to estimate the number of supporters for Candidate 3 or 4 among the additional 1,000 voters:
\[ \text{Expected number of supporters} = P(\text{Candidate 3 or 4}) \times 1,000 = 0.4 \times 1,000 = 400 \]
Answer: 400
Question 2
To find the probability of selecting an apple or an orange from the basket of fruits, we first determine the total number of fruits in the basket.
- Number of apples: 20
- Number of watermelon slices: 4
- Number of oranges: 12
Total fruits = 20 apples + 4 watermelon slices + 12 oranges = 36 fruits
Now, we calculate the total number of apples and oranges:
Total apples and oranges = 20 + 12 = 32
The probability \( P \) of selecting an apple or an orange is given by:
\[ P(\text{apple or orange}) = \frac{\text{Total apples and oranges}}{\text{Total fruits}} = \frac{32}{36} \]
We can simplify the fraction:
\[ \frac{32}{36} = \frac{8}{9} \quad (\text{divide both numerator and denominator by 4}) \]
Answer: \(\frac{8}{9}\)
Question 3
To find the probability of selecting an employee who eats breakfast or lunch, we need to use the principle of inclusion-exclusion since some employees eat both meals.
Let:
- \( A \) = Number of employees who eat breakfast = 20
- \( B \) = Number of employees who eat lunch = 50
- \( C \) = Number of employees who eat both breakfast and lunch = 15
Using the inclusion-exclusion principle, the total number of employees who eat either breakfast or lunch is calculated as:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Where \( P(A) \) is the probability of choosing someone who eats breakfast, \( P(B) \) for lunch, and \( P(A \cap B) \) for both.
Calculating \( P(A \cup B) \):
Total who eat breakfast or lunch: \[ \text{Total} = A + B - C = 20 + 50 - 15 = 55 \]
Now, the probability \( P \):
\[ P(\text{Breakfast or Lunch}) = \frac{\text{Total eating Breakfast or Lunch}}{\text{Total number of employees}} = \frac{55}{80} \]
We can simplify:
\[ \frac{55}{80} = \frac{11}{16} \quad (\text{divide both by 5}) \]
Answer: \(\frac{11}{16}\)