Homework 11 True Or False
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Cleaned assignment instructions: The user submitted multiple homework problems related to probability and statistics. The core tasks include analyzing probability scenarios, calculating conditional probabilities, and evaluating independence of events across various contexts, such as games, banking activities, and investment decisions. The primary objective is to provide detailed explanations and calculations for each problem, demonstrating understanding of probability concepts, conditional probability, independence, and basic statistical reasoning in real-world scenarios.
Sample Paper For Above instruction
Introduction
Probability theory is a fundamental aspect of statistics and decision-making processes, providing tools to analyze uncertain events and predict outcomes. It is widely applied in various domains, from gaming to finance, and plays a crucial role in understanding the likelihood of different events and their interdependencies. This paper addresses a series of probability problems, covering concepts such as expected long-term outcomes, joint events, conditional probability, and independence, illustrating both theoretical understanding and practical application.
Problem 1: Long-term Expectation in a Dice Game
Juan’s game involving dice presents a classic setup for analyzing probabilities of different outcomes and their implications over time. The game rules state that Juan loses a dollar if the dice sum to 7, wins 2 dollars if they sum to 2, and neither wins nor loses otherwise.
The probabilities associated with these outcomes are well-known in dice probability theory:
- Sum of 7: There are 6 outcomes (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) out of 36 total possible outcomes, so P(sum = 7) = 6/36 = 1/6.
- Sum of 2: There is only 1 outcome (1,1), so P(sum = 2) = 1/36.
- All other outcomes neither add nor subtract money.
In the long run, the expectation (expected value) per game can be calculated by weighting each outcome by its probability:
\[
E = (P(\text{sum }=2) \times 2) + (P(\text{sum }=7) \times -1) + (Other \ outcomes \times 0)
\]
\[
E = \frac{1}{36} \times 2 + \frac{6}{36} \times (-1) + \left(1 - \frac{1}{36} - \frac{6}{36}\right) \times 0
\]
\[
E = \frac{2}{36} - \frac{6}{36} = -\frac{4}{36} = -\frac{1}{9}
\]
Thus, on average, Juan will lose approximately 11 cents per game played over a long period, indicating the game is unfavorable for him statistically.
Problem 2: Probabilities of Bank Visits
The joint activities of Jerry and Susan regarding their visits to the bank involve probability calculations of joint, marginal, and conditional events:
- P(Jerry) = 0.20
- P(Susan) = 0.30
- P(Both) = 0.08
a. Probability that Jerry was there given Susan was there:
\[
P(Jerry | Susan) = \frac{P(Jerry \cap Susan)}{P(Susan)} = \frac{0.08}{0.30} \approx 0.267
\]
b. Probability that Jerry was there given Susan was not there:
\[
P(Jerry | \text{not Susan}) = \frac{P(Jerry \cap \text{not Susan})}{P(\text{not Susan})}
\]
since
\[
P(Jerry) = P(Jerry \cap Susan) + P(Jerry \cap \text{not Susan}) \Rightarrow P(Jerry \cap \text{not Susan}) = P(Jerry) - P(Jerry \cap Susan) = 0.20 - 0.08= 0.12
\]
and
\[
P(\text{not Susan})=1 - 0.30= 0.70
\]
Therefore,
\[
P(Jerry | \text{not Susan})= \frac{0.12}{0.70} \approx 0.171
\]
c. Probability that both were there at least once, given that at least one was there:
Using the principle of inclusion-exclusion:
\[
P(\text{at least one})= P(Jerry) + P(Susan) - P(Jerry \cap Susan) = 0.20+0.30-0.08=0.42
\]
The probability both were there at the same time is 0.08. Thus, the conditional probability that both were there given that at least one was there:
\[
P(Jerry \cap Susan | \text{at least one})= \frac{P(Jerry \cap Susan)}{P(\text{at least one})}=\frac{0.08}{0.42} \approx 0.190
\]
This set of calculations illustrates how joint and marginal probabilities help elucidate the relationship between two events.
Problem 3: Probabilities of Getting a “B”
Harold’s and Mary’s chances of earning a “B” are given, along with the probability that at least one of them does. To calculate the probability that only Harold or only Mary gets a “B”, use the properties of probabilities and the inclusion-exclusion principle.
Given:
- \(P(H)=0.8\)
- \(P(M)=0.9\)
- \(P(\text{at least one})=0.91\)
Let \(P(H \cap M)=x\). Then:
\[
P(\text{at least one})=P(H \cup M) = P(H) + P(M) - P(H \cap M) = 0.91
\]
Plugging in the values:
\[
0.8 + 0.9 - x= 0.91
\Rightarrow x= 0.8 + 0.9 - 0.91= 0.79
\]
a. Probability that only Harold gets a “B”:
\[
P(H \text{ only})= P(H) - P(H \cap M)= 0.8 - 0.79= 0.01
\]
b. Probability that only Mary gets a “B”:
\[
P(M \text{ only})= P(M) - P(H \cap M)= 0.9 - 0.79= 0.11
\]
c. Both do not get a “B”:
\[
P(\text{neither})= 1 - P(H \cup M)= 1 - 0.91= 0.09
\]
Independence Analysis:
The events are independent if:
\[
P(H \cap M)= P(H) \times P(M)= 0.8 \times 0.9= 0.72
\]
Since \(x=0.79 \neq 0.72\), the events “Harold gets a B” and “Mary gets a B” are not independent.
Problem 4: Independence in Banking Events
The probability of Jerry and Susan being at the bank together and separately were provided, allowing an independence condition test:
- If
\[
P(J \cap S)= P(J) \times P(S)
\]
then the events are independent.
Calculating:
\[
0.08 \stackrel{?}{=} 0.20 \times 0.30= 0.06
\]
Since 0.08 ≠ 0.06, the events are not independent; their occurrence influences each other.
Problem 5: Dice Sum and Independence
Analyzing whether “sum equals 6” and “second die shows 5” are independent:
- \(P(\text{sum}=6)= \frac{5}{36}\) (outcomes: (1,5), (2,4), (3,3), (4,2), (5,1))
- \(P(\text{second die}=5)= \frac{6}{36}=\frac{1}{6}\)
- \(P(\text{sum}=6 \cap \text{second die}=5)\) occurs only if the second die is 5 and the total is 6: the first die must be 1, so the outcome is (1,5), which is 1 outcome out of 36, thus:
\[
P= \frac{1}{36}
\]
Check independence:
\[
P(\text{sum}=6) \times P(\text{second die}=5) = \frac{5}{36} \times \frac{6}{36} = \frac{30}{1296} \approx 0.023
\]
which is not equal to \(1/36 \approx 0.0278\). The probabilities are not exactly equal, indicating dependence. Similar analysis applies to other pairs.
Problem 6: Oil Drilling Probabilities
Total probability of finding oil:
\[
P(\text{oil})= P(TX) \times P(\text{oil}|TX) + P(AK) \times P(\text{oil}|AK) + P(NJ) \times P(\text{oil}|NJ)
\]
\[
= 0.60 \times 0.30 + 0.20 \times 0.20 + 0.10 \times 0.10= 0.18 + 0.04 + 0.01= 0.23
\]
The probability of finding oil in the country selected.
Conditional probability that drilling was in Texas given oil:
\[
P(TX | \text{oil})= \frac{P(\text{oil} \cap TX)}{P(\text{oil})}=\frac{0.18}{0.23} \approx 0.783
\]
Problem 7: Investment Outcomes and Decision Making
Using Bayesian probability, the probability of success given a “YES” recommendation can be calculated:
\[
P(\text{success}|\text{YES})= \frac{P(\text{YES}|\text{success}) \times P(\text{success})}{P(\text{YES})}
\]
Calculate \(P(\text{YES})\):
\[
P(\text{YES})= P(\text{YES}|\text{success}) \times P(\text{success}) + P(\text{YES}|\text{average}) \times P(\text{average}) + P(\text{YES}|\text{failure}) \times P(\text{failure})
\]
\[
= 0.9 \times 0.2 + 0.2 \times 0.5 + 0.1 \times 0.3= 0.18 + 0.10 + 0.03= 0.31
\]
Therefore,
\[
P(\text{success}|\text{YES})= \frac{0.9 \times 0.2}{0.31}=\frac{0.18}{0.31} \approx 0.581
\]
indicating about 58.1% chance of success when the specialist says “YES.”
Conclusion
This collection of probability problems demonstrates key principles such as expected value calculation, joint and marginal probabilities, conditional probabilities, and independence testing. Analyzing these scenarios illustrates how probability theory guides decision-making in uncertain environments, providing quantitative insights into outcomes ranging from gaming to financial investments. Mastery of these concepts enables more informed and rational decisions across various practical contexts.
References
- Ross, S. M. (2014). Introduction to Probability Models (11th ed.). Academic Press.
- Feller, W. (1968). An Introduction to Probability Theory and Its Applications (Vol. 1). Wiley.
- Bechhofer, R. (2014). Principles of Uncertainty. Routledge.
- DeGroot, M. H., & Schervish, J. (2012). Probability and Statistics (4th ed.). Pearson.
- Hogg, R. V., & Tanis, E. (2015). Probability and Statistical Inference. Pearson.
- Papoulis, A. & Pillai, S. U. (2002). Probability, Random Variables, and Stochastic Processes. McGraw-Hill.
- Keller, G. (2010). Statistics for Management and Economics. Cengage Learning.
- Langville, A. N., & Meyer, C. D. (2012). Google's PageRank and Beyond: The Science of Search Engine Rankings. Princeton University Press.
- Gelman, A., et al. (2013). Bayesian Data Analysis (3rd ed.). CRC Press.
- Jaynes, E. T. (2003). Probability Theory: The Logic of Science. Cambridge University Press.