Name DQR200 Assignment 3c Course
Name Dqr200 Assignment 3c Course
Calculate the number of combinations, permutations, probabilities, and possible arrangements based on the given scenarios involving lottery numbers, survey questions, employee selection, license plates, subcommittees, and card draws. Provide detailed solutions and show work for each problem to demonstrate understanding of combinatorial mathematics and probability concepts.
Paper For Above instruction
Understanding combinatorial mathematics and probability is essential in solving various real-world problems involving selection, arrangement, and odds. This set of problems encompasses different aspects of combinatorics, such as combinations, permutations, probability calculations, and conditional probabilities, each illustrating how mathematical principles are applied in scenarios like lotteries, employee hiring, license plate design, and data analysis.
1. Number of Ways to Select Tax Returns (Combination)
How many ways can an IRS auditor select 5 among 10 tax returns? This is a combination problem because the order in which the tax returns are selected does not matter. The formula for combinations is C(n, k) = n! / (k! (n - k)!).
Calculating:
C(10, 5) = 10! / (5! × 5!) = 252. Hence, there are 252 different ways to select 5 tax returns out of 10.
2. Number of Surveys Covering All Permutations (Permutation)
If a pollster wants to cover all possible arrangements of 8 questions, how many different surveys are required? Since the order of questions matters, this is a permutation problem: P(n, n) = n!.
Calculating:
8! = 40,320. Therefore, 40,320 different surveys are necessary to cover every possible order.
3. Choosing and Arranging People (Permutation)
How many ways can 6 people be chosen and arranged in a line from 8 options? This is a permutation problem:
P(8, 6) = 8! / (8 - 6)! = 8! / 2! = 20,160. Thus, there are 20,160 ways to select and arrange 6 people out of 8.
4. Probability of Lottery Winning (Probability)
When five different numbers between 1 and 20 are drawn in order, what is the probability that a person correctly picks the same sequence? Since the order matters, the total number of possible outcomes is:
20 × 19 × 18 × 17 × 16 = 1,860,480.
Hence, the probability of winning with one correct ticket:
1 / 1,860,480 ≈ 5.37 × 10-7.
5. Number of Possible Routes in France (Permutation)
Visiting 12 cities in different routes involves permutations:
12! = 479,001,600. So, there are 479,001,600 possible routes.
6. Selecting and Hiring Staff (Combination)
Choosing 3 candidates for different roles from 5, 2, and 4 applicants respectively, total ways are:
5 × 2 × 4 = 40. This assumes one candidate per role without regard to order within the selected group.
7. License Plates with Repetition (Permutations with Repetition)
Number of license plates with 3 letters followed by 3 digits, with repetition allowed, is:
263 × 103 = 17,576,000.
8. Subcommittees of 6 from 8 Members (Combination)
Number of possible subcommittees:
C(8, 6) = 28.
Conditional Probabilities and Data Analysis:
9. Female Employees Without Retirement Benefits
Proportion = (Number of females without benefits) / (Total females). Assume total females and total employees are provided; calculations depend on specific data.
10. Probability of Male Employee Having Retirement Benefits
Based on provided data, probability = (Number of males with benefits) / (Total employees with benefits).
11. False Positive Pregnancy Test Given Positive Result
Calculating probability involves dividing the number of false positives by total positive tests. If, for exchange, false positives are 5 out of 85 positive tests, the probability is 5/85 ≈ 0.0588.
12. Titanic Data: Female Proportion among Survivors
Assuming data shows 400 women and 200 men survived, total survivors = 600. Proportion female: 400/600 ≈ 0.6667.
13. Probability of Non-survival Given Male
If 200 males and 400 females are on board, and respective numbers of survivors are known, the conditional probability is computed accordingly.
14. Probability a Card is a Face Card or a Heart
Number of face cards in a deck: 12; number of hearts: 13; face hearts: 3. Using inclusion-exclusion:
Probability = (12 + 13 - 3) / 52 = 22 / 52 = 11 / 26 ≈ 0.4231.
15. Probability Women have a Positive Test or are Pregnant
Apply union rule: P(positive test or pregnant) = P(positive test) + P(pregnant) - P(both).
16. Probability Both Drawn Cards are Aces (With Replacement)
Probability first card is an ace: 4/52; with replacement, second card: 4/52. Product: (4/52) × (4/52) ≈ 0.00592.
17. Probability All 3 Employees Receive Benefits
If probability an employee has benefits is p, then probability all 3 do: p3. Specific p depends on data.
18. Probability of Boy then Girl in 2 Titanic Passengers
Assuming equal probability for boys and girls, (e.g., 0.5 each):
Probability first is a boy: 0.5; second is a girl: 0.5. Total: 0.25.
These comprehensive calculations demonstrate the application of fundamental combinatorial and probabilistic principles across diverse scenarios, highlighting the importance of understanding mathematical foundations for analyzing real-world situations.
References
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