MTH 165 Lund Chapter 5 Take Home Quiz Covering 51 52 5
Mth 165 Lundchapter 5 Take Home Quiz Covering 51 52 5
MTH 165 Lund Chapter 5 Take-home Quiz Covering 5.1, 5.2, 5.3, 5.4 NO TUTORING CENTER HELP. No credit will be given without supporting work. A quiz consisting of multiple pages must be stapled. Work must be legible and completed on a separate piece of paper. All answers should include probability notation!
Round all answers to 4 decimal places.
Paper For Above instruction
This assignment involves a comprehensive application of probability concepts, including conditional probability, independence, probability distributions, and combinatorial probability. The questions are based on real-world contexts such as automobile accidents, die rolls, marble draws, loan default risks, and student punctuality. Each problem requires careful formulation of probability models, calculations with probability notation, and interpretation of results. Supporting work, including formulas, probability expressions, and reasoning, is essential for full credit.
Question 1: Automobile Accident Data
A sample of 500 people, all driving approximately 10,000 miles annually, is classified according to age and accident history over the past three years. The classification involves two factors: age (
a. Find the probability that a randomly selected person is over 40 and has had more than 1 accident.
b. Find the probability that the person is either over 40 or has had more than 1 accident (or both).
c. Find the probability that the person is over 40 given that they have had more than 1 accident.
d. If the selected person is over 40, what is the probability that they have had more than 1 accident?
e. Find the probability that the person is over 40 given they have had at most 1 accident.
f. Based on the data, evaluate whether the events "Being over 40 years old" and "Getting in more than 1 accident" are independent. Justify your answer with appropriate probability calculations.
Question 2: Loaded Die Probability Model
A six-sided die is loaded such that the probability outcomes are proportional to unknown value p, with P(1)=P(2)=P(3)=P(4)=P(5)=P(6). Determine p for the probabilities to sum to 1 and specify each probability.
a. Find the value of p that makes this a valid probability distribution and state the exact probability for each face.
b. Compare two events, such as rolling a 1 or rolling a 6, and identify which is more likely based on the calculated probabilities.
Question 3: Marble Selection from a Bag
A bag contains 6 green, 7 yellow, and 3 purple marbles. Two marbles are drawn sequentially without replacement.
a. Calculate the probability that the first marble is green and the second is yellow.
b. Find the probability of drawing no green marbles in two picks.
c. Determine the probability of selecting exactly one purple marble and one non-purple marble, considering all possible cases.
Question 4: Bank Loan Default and Risk Analysis
A bank categorizes loans as "high risk" (20%) or "low risk," with 5% overall default rate. It is known that if a loan defaults, there is a 40% chance it was issued to a high-risk borrower.
a. Compute the probability that a randomly selected loan is in default and issued to a high-risk borrower.
b. Find the probability that a loan will default given it was issued to a high-risk borrower.
c. Calculate the probability that a loan is either in default or issued to a high-risk borrower, or both.
d. If a loan is issued to a low-risk borrower, what is the probability it will default?
Question 5: Student Punctuality
Maxine attends school 4 days a week. Each day, she can arrive on time or be late, with the probability of lateness on any day being 0.08. The lateness on different days is independent.
a. Determine the probability that she is on time for the first two days and late for the next two days.
b. Calculate the probability that she is late on at least one day during the four-day week.
Solution to Above instruction
Question 1: Automobile Accident Data
Let’s define the subgroups and relevant probabilities based on the data. Assume the total sample size is n = 500. We denote the following events:
- O: Person is over 40
- Y: Person is under 40
- M: Person has had more than 1 accident
- N: Person has had at most 1 accident
Without specific frequencies provided, we describe the calculations in terms of the given probabilities and known totals.
- The probability that a randomly selected person is over 40 and had more than 1 accident is:
\[ P(O \cap M) = \frac{\text{Number of people over 40 with >1 accidents}}{500} \]
Since explicit counts are not provided, assume the joint probability is identified or estimated from data or contingency tables. The calculation would resemble:
\[ P(O \cap M) = P(O) \times P(M|O) \]
- The probability that a person is either over 40 or has had more than 1 accident is:
\[ P(O \cup M) = P(O) + P(M) - P(O \cap M) \]
- Conditional probability that a person is over 40 given that they had more than 1 accident:
\[ P(O|M) = \frac{P(O \cap M)}{P(M)} \]
- Probability that a person over 40 has had more than 1 accident:
\[ P(M|O) = \frac{P(O \cap M)}{P(O)} \]
- Probability that a person over 40 has had at most 1 accident:
\[ P(N|O) = 1 - P(M|O) \]
- Independence check: To determine if the events are independent:
\[ P(O \cap M) \stackrel{?}{=} P(O) \times P(M) \]
If equality holds, the events are independent; otherwise, they are dependent.
Question 2: Loaded Die Probability Model
The probability of each outcome is proportional to an unknown p, with the sum of probabilities equal to 1.
The probabilities are: P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = p.
Sum of probabilities:
\[ 6p = 1 \implies p = \frac{1}{6} \]
Thus, each face has a probability of \(\frac{1}{6}\), making it a fair die.
Comparison of events such as rolling a 1 versus rolling a 6 shows both have equal probability of \(\frac{1}{6}\); hence, they are equally likely.
Question 3: Marble Selection from a Bag
Total marbles: 6 green, 7 yellow, 3 purple; total = 16.
- Probability first green, second yellow:
\[ P(G_1 \cap Y_2) = P(G_1) \times P(Y_2 | G_1) = \frac{6}{16} \times \frac{7}{15} = \frac{6}{16} \times \frac{7}{15} = \frac{42}{240} = \frac{7}{40} \]
- No green in two picks:
\[ P(\text{no green}) = P(\text{both yellow or purple}) = \frac{10}{16} \times \frac{9}{15} = \frac{10}{16} \times \frac{9}{15} = \frac{90}{240} = \frac{3}{8} \]
- Exactly one purple and one non-purple:
There are two arrangements: Purple then non-purple, or non-purple then purple.
\[ P(\text{Purple then non-purple}) = \frac{3}{16} \times \frac{13}{15} = \frac{39}{240} = \frac{13}{80} \]
\[ P(\text{Non-purple then purple}) = \frac{13}{16} \times \frac{3}{15} = \frac{39}{240} = \frac{13}{80} \]
Adding both:
\[ P = 2 \times \frac{13}{80} = \frac{26}{80} = \frac{13}{40} \]
Question 4: Bank Loan Default and Risk Analysis
Given Data:
- \( P(H) = 0.20 \) (High risk)
- \( P(L) = 0.80 \) (Low risk)
- \( P(D) = 0.05 \) (Default rate)
- Conditional probability: \( P(H | D) = 0.40 \)
Using Bayes' theorem and joint probability calculations.
- Probability default and high risk:
\[ P(D \cap H) = P(H) \times P(D | H) \]
From \( P(H|D) \):
\[ P(H \cap D) = P(D) \times P(H | D) = 0.05 \times 0.40 = 0.02 \]
- Probability default given high risk:
\[ P(D | H) = \frac{P(H \cap D)}{P(H)} = \frac{0.02}{0.20} = 0.10 \]
- Probability that a loan is either in default or issued to high risk:
\[ P(D \cup H) = P(D) + P(H) - P(H \cap D) = 0.05 + 0.20 - 0.02 = 0.23 \]
- Probability that a loan issued to a low-risk borrower defaults:
\[ P(D | L) = \frac{P(D) - P(H \cap D)}{P(L)} = \frac{0.05 - 0.02}{0.80} = \frac{0.03}{0.80} = 0.0375 \]
Question 5: Maxine’s Punctuality
Probability she is late on any given day: \( p = 0.08 \).
In the case of four independent days, the probability she is on time each day is \( (1 - p)^4 \).
- Probability she is on time first two days and late for next two days:
\[ P(\text{On time 1 and 2, late 3 and 4}) = (1 - 0.08)^2 \times 0.08^2 = 0.92^2 \times 0.08^2 = (0.8464) \times (0.0064) \approx 0.0054 \]
- Probability she is late on at least one day:
\[ P(\text{at least one late}) = 1 - P(\text{all on time}) = 1 - (1 - 0.08)^4 = 1 - 0.92^4 \approx 1 - 0.7164 = 0.2836 \]
References
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- Devore, J. L. (2015). Probability and Statistics for Engineering and Science. Cengage Learning.
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- Feller, W. (1968). An Introduction to Probability Theory and Its Applications. Wiley.
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