John Madean's Coin Toss Experiment

1 John Madean Experiment By Tossing Three Fair Coins Faircoin Has T

Analyze multiple probability and statistics problems involving experiments with coins, vehicle ownership among students, committee selection from employees, normal distribution calculations, and binomial distribution identification, as well as substance abuse testing in construction companies.

Paper For Above instruction

In this paper, we explore a series of problems that involve fundamental concepts in probability, combinatorics, and statistics. These problems encompass experiments with coin tosses, probability calculations based on vehicle ownership data, combinatorial analysis for committee selection, properties of normal distribution, identification of binomial distributions, and applications of probability in testing for substance abuse among companies. Each section will analyze the specific problem, apply relevant mathematical principles, and interpret the results with appropriate statistical reasoning.

Problem 1: Coin Toss Experiment

John conducted an experiment where he tossed three fair coins. To analyze this, we begin by identifying the sample space—the set of all possible outcomes. Since each coin has two equally likely outcomes—heads (H) or tails (T)—the number of total possible outcomes when tossing three coins is \(2^3 = 8\). These outcomes include:

  • HHH
  • H T H
  • H T T
  • T H H
  • T H T
  • T T H
  • T T T

Next, the probability of getting three tails (TTT) is calculated by noting that each outcome is equally likely and independent. The probability of each specific outcome, such as TTT, is \(\frac{1}{8}\).

Similarly, the probability of exactly two tails involves outcomes with precisely two T's: H T T, T H T, and T T H. There are 3 such outcomes, so the probability is \(\frac{3}{8}\).

Finally, the probability of at least one tail is computed by considering all outcomes except for the scenario with zero tails (which is only H H H). Since only H H H has no tails, the probability of zero tails is \(\frac{1}{8}\). Therefore, the probability of at least one tail is:

\(\)1 - P(no tails) = 1 - \(\frac{1}{8}\) = \(\frac{7}{8}\).

Problem 2: Vehicle Ownership Among UMUC Students

Given data indicate that 70% of students own a car, 50% own a bike, and 40% own both. To analyze these, we utilize a Venn diagram to visualize the relationships between these sets. The probability that a randomly chosen student owns at least one of the two vehicles is calculated via the principle of inclusion-exclusion:

P(owning a car or bike) = P(car) + P(bike) - P(both) = 0.70 + 0.50 - 0.40 = 0.80.

Hence, there is an 80% chance that a randomly chosen student owns at least one of these vehicles.

The probability that a student owns a bike, regardless of other possessions, is directly given as 50% or 0.50.

Problem 3: Committee Selection from Employees

In a division with 12 employees (8 males, 4 females), several combinatorial selections are examined. The total number of ways to form a 6-member committee from 12 employees can be determined using combinations:

\(\)C(12,6) = \(\frac{12!}{6! \times 6!}\) = 924.

The probability that a randomly selected committee contains all males (8 males, choosing 6 from these 8):

\(\)C(8,6) / C(12,6) = \(\frac{28}{924}\) = approximately 0.0303.

For the committee with a female president and two male members: choose 1 female from 4 and 2 males from 8, then select additional members to fulfill the total of 6, ensuring roles are specific. The number of such committees involves factorial calculations considering the distinct roles, which in this case, yields:

\(\)Number of committees = C(4,1) \times C(8,2) \times C(9,3), assuming remaining members are chosen from the remaining employees, but specific constraints should be clarified for precise calculation.

Problem 4: Normal Distribution Calculations

For a normally distributed variable \(X\) with mean 12 and standard deviation 4:

a) The probability that \(X\) exceeds 10 is calculated by finding the Z-score:

\(Z = \frac{10 - 12}{4} = -0.5\).

Using standard normal distribution tables, \(P(Z > -0.5)\) ≈ 0.6915.

b) For \(X

c) For the interval [10, 14.5], calculate Z-scores: Z for 14.5 is \(\frac{14.5 - 12}{4} = 0.625\). The probability is \(P(Z

d) To find \(a\) such that \(P(X

e) To find \(b\) such that \(P(X > b) = 0.15\), Z = \(-1.036\), so \(b = 12 + (-1.036) \times 4 = 7.856.\)

Binomial Distribution Identification

Determining which of the listed scenarios is not a binomial distribution involves understanding binomial assumptions: fixed number of independent trials, two possible outcomes, constant probability. Scenario 3, "Rolling a fair die 10 times and keeping track of all numbers rolled," does not fit because it involves multiple outcomes beyond success/failure, thus violating the binomial criteria.

Substance Abuse Testing in Construction Companies

Out of companies engaged in highway or bridge construction, 80% test employees for substance abuse. From a sample of 12 companies:

a) The probability exactly 7 out of 12 test for substance abuse is:

\(P = C(12,7) \times (0.8)^7 \times (0.2)^5\) ≈ 0.214.

b) The probability at least half (6 or more) test for substance abuse involves summing probabilities from k=6 to k=12, which can be computed using binomial formulas, resulting in approximately 0.953.

c) The mean number of companies testing is \(12 \times 0.8 = 9.6\), and the standard deviation is \(\sqrt{12 \times 0.8 \times 0.2} \approx 1.55.\)

d) Using the range rule of thumb: the usual range is approximately mean ± 2 standard deviations, i.e., 9.6 ± 3.10, or roughly 6.5 to 12.7.

Conclusion

This comprehensive analysis demonstrates core statistical concepts such as probability calculations, combinatorial analysis, normal distribution applications, and understanding the properties of binomial distributions. Each problem illustrates practical applications of theoretical principles, essential for proficiency in statistics and probability analysis.

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