Mutually Inclusive and Exclusive Events – Portfolio
Question 1: There are four presidential candidates in a country with the following survey results:
- Candidate 1: 250
- Candidate 2: 1,250
- Candidate 3: 450
- Candidate 4: 550
Probability of Voting for Candidate 1 or 2:
Total number of voters surveyed: 2,500
Number of voters for Candidate 1 or Candidate 2:
- Voters for Candidate 1 = 250
- Voters for Candidate 2 = 1,250
Using the formula for probability of mutually exclusive events: \[ P(A \text{ or } B) = P(A) + P(B) \]
Calculating the probability: \[ P(1 \text{ or } 2) = \frac{250}{2500} + \frac{1250}{2500} = \frac{1500}{2500} = 0.6 \]
Thus, the probability of randomly selecting a person who will vote for Candidate 1 or 2 is 0.6 or 60%.
If 1,000 more voters are surveyed, how many will likely prefer Candidate 3 or 4?
The proportion of current voters for Candidate 3 and 4:
- Candidate 3: 450
- Candidate 4: 550
- Total for Candidate 3 and 4 = 450 + 550 = 1,000
Thus, the proportion is \( \frac{1000}{2500} = 0.4 \)
If 1,000 more voters are surveyed: \[ 0.4 \times 1000 = 400 \] So, it can be expected that approximately 400 voters will prefer Candidate 3 or 4 among the new voters surveyed.
Question 2: A basket of fruits has 20 apples, 4 watermelon slices, and 12 oranges.
Probability of Selecting an Apple or an Orange:
Total number of fruits: \[ 20 + 4 + 12 = 36 \]
Number of apples or oranges: \[ 20 + 12 = 32 \]
Using the probability formula: \[ P(A \text{ or } O) = \frac{32}{36} = \frac{8}{9} \approx 0.8889 \]
Thus, the probability of selecting an apple or an orange is approximately 0.889 or 88.89%.
Question 3: In a survey of 80 employees:
- Employees that eat breakfast: 20
- Employees that eat lunch: 50
- Employees that eat both breakfast and lunch: 15
Probability of Selecting an Employee who Eats Breakfast or Lunch:
Using the formula for the probability: \[ P(B \text{ or } L) = P(B) + P(L) - P(B \text{ and } L) \]
Calculating the individual probabilities: \[ P(B) = \frac{20}{80}, \quad P(L) = \frac{50}{80}, \quad P(B \text{ and } L) = \frac{15}{80} \]
Now substituting into the formula: \[ P(B \text{ or } L) = \frac{20}{80} + \frac{50}{80} - \frac{15}{80} = \frac{55}{80} = 0.6875 \]
Thus, the probability of randomly selecting an employee who eats breakfast or lunch is 0.6875 or 68.75%.
Dependent and Independent Events – Portfolio
Question 4: Using the letters of the word CLEAN, simulate a spinner to generate two letters (with repetition).
Table summarizing 30 trials: | Trial | Letter 1 | Letter 2 | Combination | |-------|----------|----------|---------------| | 1 | C | E | CE | | 2 | A | N | AN | | 3 | L | E | LE | | 4 | C | C | CC | | ... | ... | ... | ... | | 30 | E | A | EA |
(Below you will input the results from your trials and tally the combinations.)
Probability of Spinning Letters C and E:
- Count the occurrences of combinations with C and E.
- For example, if CE appears 4 times and EC appears 3 times, that’s 7 out of 30.
Calculating the theoretical probability:
- There are a total of \(5 \times 5 = 25\) combinations.
- Favorable outcomes for (C, E) or (E, C) = 2.
So, the theoretical probability is: \[ P(C \text{ and } E) = \frac{2}{25} = 0.08 \]
Comparison: Compare the experimental probability to the theoretical probability results.
Question 5: For a private subdivision with 6 house models (A, B, C, D, E, F) and options of garage or carport.
Simulation Summary: Generate random combinations to see how many select B with a car garage. For instance: | Count | Garage | Carport | |-------|--------|---------| | 1 | B | - | | 2 | C | - | | 3 | B | - | |... | ... | ... |
After running the simulation for a sufficient number of times:
If model B with a car garage appears 12 times out of 60 trials: \[ P(B \text{ with garage}) = \frac{12}{60} = 0.2 \]
Thus, the probability that a client will choose model B with a car garage is 0.2 or 20%.
Make sure to save your worksheet with your finalized answers for submission!