Introduction To Mathematical Statistics In The Game
Introduction To Mathematical Statistics1 In The Game Three Card Mont
Introduction to Mathematical Statistics 1. In the game ‘Three Card Monte’ there are 3 cards, one of them a queen. The three cards are placed face down on a table. The dealer shuffles the cards quickly while they are face down on the table. The player selects a card; the goal is to find the queen.
A. What is the probability you DO NOT choose the queen with your initial guess?
B. Before showing you the card you have chosen, the dealer flips one of the cards. It is not the queen. The dealer then offers you the option to keep the card you have chosen or take the remaining card. If you decide to pick the remaining card, what is the probability you choose the queen?
C. What is the probability you choose the queen with your initial guess?
I write each of the letters ‘P’, ‘R’, ‘I’, ‘D’, ‘E’ on separate pieces of paper and put these into a hat. I close my eyes and begin picking the pieces of paper from the hat and reading off the letters written on the pieces of paper.
1. If I do NOT put the pieces of paper back into the hat after each choice and pick 5 pieces of paper, what is the probability that the letters come out exactly in the order ‘PRIDE’?
2. If I put the pieces of paper back into the hat after each choice and pick 5 pieces of paper, then what is the probability that the letters come out exactly in the order ‘PRIDE’?
3. If I put the pieces of paper back into the hat after each choice and pick 5 pieces of paper, then what is the probability that the letters picked can be rearranged to spell ‘PRIDE’?
4. If I put the pieces of paper back into the hat after each choice and pick 5 pieces of paper, then what is the probability that there are exactly two ‘P’ letters among the 5 letters picked?
5. If I put the pieces of paper back into the hat after each choice and continue to pick letters until the first ‘P’ occurs, what is the probability the first ‘P’ occurs as the 4th letter picked?
6. Suppose 1% of people have a certain genetic mutation. 90% of tests for the gene mutation detect it in those who have it while 9.6% of the tests falsely detect it in those who do not have it.
7. If a person is chosen at random, what is the probability the person tests positive for the genetic mutation? (Note: This involves computing total probability based on prevalence and test accuracy.)
8. If a randomly chosen person tests positive for the genetic mutation, what is the probability they actually have it? (Note: This involves applying Bayes' theorem.)
StatCo manufactures and sells desktop computers and monitors. Suppose that out of those customers who visit StatCo Computer Stores, 60% buy desktops, 60% buy monitors, and 90% buy either a monitor or a desktop or both.
9. What is the probability that a customer buys a desktop and a monitor?
10. What is the probability that a customer buys a desktop, given that they bought a monitor?
11. What is the probability that a customer buys a desktop, given that they did not buy a monitor?
A study was conducted to determine if a medicine helps to shorten the length of a cold. 200 subjects were recruited who had suffered from a cold and either took the medicine or not. The length of times in days the subjects suffered from the cold was noted. The following table contains the results:
| Medicine? | Cold Length |
|---|---|
| Yes | 1-3 Days |
| No | 1-3 Days |
12. If a subject is chosen at random, what is the probability the subject both took the medicine and had the cold for 1-3 days?
13. If a subject is chosen at random, what is the probability the subject took the medicine given that the subject’s cold lasted for 1-3 days?
14. Are the events ‘subject took the medicine’ and ‘cold lasted for 1-3 days’ independent or dependent? Why or why not?
Paper For Above instruction
Mathematical statistics offers vital tools to analyze probability scenarios, such as those presented in games like Three Card Monte, and in real-world situations like medical testing, consumer behavior, and genetic screening. This paper explores various probability concepts through these scenarios, highlighting the application of fundamental principles such as conditional probability, Bayes' theorem, and independence.
Analysis of the Three Card Monte Game
The initial game involving three cards with one queen illustrates basic probability principles. When choosing a card at random, the probability of not selecting the queen initially is straightforward. Since there are three cards with one Queen, the probability of selecting any other card is 2/3. Therefore, the probability of not choosing the queen initially is 2/3 (or approximately 66.67%) (Feller, 1968).
After the dealer reveals a non-queen card, the probability dynamics shift significantly. Keeping the initial card yields the initial probability, but switching to the remaining card has a conditional probability. If the player switches, the probability of winning increases because the initial choice had a one-third chance of being the queen, and the revealed card confirms the remaining card is more likely to be the queen. This scenario aligns with the Monty Hall problem, where switching improves chances to 2/3 (Hu, 1997).
Hence, the probability of choosing the queen initially remains 1/3, illustrating the importance of understanding conditional probability and the benefits of switching in such games.
Probability of Specific Letter Arrangements
Next, the problem involving drawing letters from a hat without or with replacement demonstrates permutation and combination principles. When drawing 5 letters without replacement, the probability of a specific order (e.g., PRIDE) is calculated by the product of decreasing probabilities: 1/5 for the first letter, 1/4 for the second, etc., resulting in (1/120) or approximately 0.00833 (Williams, 2012).
With replacement, each draw is independent, and the probability of the exact order PRIDE remains 1/120, as each position has a 1/5 chance. The likelihood that the set of letters can be rearranged to spell PRIDE involves combinatorics, and the probability of exactly two P's among five draws follows the binomial distribution, given by C(5, 2)(1/5)^2(4/5)^3, resulting in approximately 0.2048.
The scenario where the first P occurs as the fourth letter is modeled using geometric distribution considerations, resulting in a probability of (4/5)^3*(1/5) ≈ 0.16384, indicating that the first P is more likely to occur later in the sequence.
Genetic Testing and Bayes' Theorem
In genetic testing scenarios, understanding the probability that an individual has a mutation given a positive test involves applying Bayes' theorem. Given that 1% of the population has the mutation, tests detect it 90% of the time among carriers, but also have a 9.6% false positive rate among non-carriers. The overall probability of testing positive combines these facts:
P(Positive) = P(Positive | mutation) P(mutation) + P(Positive | no mutation) P(no mutation) = (0.9)(0.01) + (0.096)(0.99) ≈ 0.009 + 0.095 ≈ 0.104 (Mccall, 2019).
Using Bayes' theorem, the probability the person actually has the mutation given a positive test is:
P(mutation | positive) = (P(positive | mutation)*P(mutation)) / P(positive) ≈ 0.009 / 0.104 ≈ 0.0865, or 8.65% (Rothman & Greenland, 1998).
Consumer Purchase Behavior and Probability
The probabilities involving customers’ purchase decisions demonstrate the application of set theory and conditional probability. The probability that a customer buys both a desktop and a monitor can be derived by subtracting the union probability from the sum of individual probabilities:
P(desktop ∩ monitor) = P(desktop) + P(monitor) - P(desktop ∪ monitor) = 0.6 + 0.6 - 0.9 = 0.3.
The conditional probability that a customer buys a desktop given they bought a monitor is:
P(desktop | monitor) = P(desktop ∩ monitor) / P(monitor) = 0.3 / 0.6 = 0.5.
Similarly, the probability that a customer buys a desktop given they did not buy a monitor is:
P(desktop | no monitor) = (P(desktop) - P(desktop ∩ monitor)) / P(no monitor) = (0.6 - 0.3) / 0.4 = 0.3 / 0.4 = 0.75 (Agresti, 2018).
Medical Study on Cold Duration
The study assessing the effectiveness of a cold medicine utilizes contingency tables and probability calculations. The probability that a randomly chosen subject both took medicine and had a cold duration of 1-3 days can be calculated by dividing the count of such cases by the total number:
P(medicine ∩ 1-3 days) = (Number of subjects with both) / 200. Assuming the counts are available, the exact probability follows this ratio (Casella & Berger, 2002).
The probability that a subject took medicine given the cold lasted 1-3 days is obtained via conditional probability:
P(took medicine | 1-3 days) = P(took medicine ∩ 1-3 days) / P(1-3 days). This can be interpreted as assessing whether medicine influences cold duration, and whether the events are independent or dependent.
Given the structure, the dependence can be examined by comparing P(took medicine) and P(took medicine | 1-3 days). If they are equal, events are independent; otherwise, dependent. The answer hinges on the observed data, but generally, an observed difference suggests dependence, indicating that medicine usage influences cold duration (Bishop & Fienberg, 1975).
Conclusion
This exploration demonstrates the critical role of probability theory in analyzing diverse practical scenarios—from gambling games and letter arrangements to medical testing and consumer behavior. By understanding conditional probability, Bayes' theorem, and independence, analysts can interpret data more robustly, make informed decisions, and develop effective strategies. Advances in statistical methods continue to enhance these applications, underscoring the importance of statistical literacy in scientific and practical contexts.
References
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