There Are Five Problems In This Assignment For Which You Sho
There Are Five Problems In This Assignment For Which You Should Submit
There are five problems in this assignment for which you should submit solutions. Please clearly label all problems and show all work! In particular, some questions specifically call for explanations. These are important.
Paper For Above instruction
Problem 1: Probability of obtaining 2 heads and 2 tails in 4 coin flips
Pat claims that the probability is 1/16, reasoning there is only one way to get two heads and two tails in the specific order, and that the total number of possible outcomes for four flips is 16 (2^4). Sam argues that the probability is 6/16, referencing the sixth entry in the fourth row of Pascal's triangle, which corresponds to the number of ways to get 2 heads in 4 flips, with the total outcomes being 16. The correct understanding aligns with Sam's explanation because the probability of getting exactly two heads and two tails in four flips is computed by dividing the number of favorable outcomes (which is 6, as given by combinatorial calculations) by the total possible outcomes (16). Pat's calculation considers only one specific ordered outcome, which is incorrect in this context, since all arrangements of two heads and two tails are equally likely in a binomial experiment. Therefore, Sam is correct because his reasoning accurately reflects the binomial probability calculation, considering all arrangements with exactly two heads in four flips.
Problem 2: Relationship between outcomes of 5 heads and 5 tails in 10 flips, and outcomes of 4 or 5 heads in 9 flips
The number of outcomes in which we obtain 5 heads and 5 tails in 10 flips can be understood recursively: these outcomes can be partitioned based on the result of the first flip. Suppose we analyze the first flip; if it’s a head, then the remaining 9 flips must contain exactly 4 heads (since total heads are 5), which equates to the number of outcomes with 4 heads in 9 flips. Conversely, if the first flip is a tail, then the remaining 9 flips need to contain 5 heads. The total outcomes with 5 heads and 5 tails in 10 flips thus sum up from these two mutually exclusive scenarios, matching the sum of outcomes with 4 heads in 9 flips and 5 heads in 9 flips. This recursive structure reflects the binomial coefficient relationship, reinforcing why the total counts are connected in this way.
Problem 3: Coin fairness testing and probability calculations
a. When flipping a fair coin 10 times, the probability of correctly concluding the coin is fair involves the probability that the number of heads lies within the acceptable range (difference at most 2 from half the flips, i.e., between 4 and 6 heads). This encompasses summing the probabilities of getting exactly 4, 5, or 6 heads, each computed via binomial distribution: P(X=k) = C(10, k) * (1/2)^10. Calculating these, the total probability of a correct conclusion when the coin is fair is the sum of these probabilities, approximately 0.849.
b. The probability of incorrectly concluding the coin is not fair under these conditions occurs when the number of heads is outside the acceptable inference range (less than 4 or greater than 6) but the coin is actually fair. The sum of these tail probabilities, computed via binomial distribution, is approximately 0.151, representing the Type I error rate.
c. Increasing the number of flips to 12 generally improves the accuracy of the test by reducing the variability inherent in smaller samples, thereby increasing the probability of correct conclusions when the coin is fair. Specifically, the larger sample size reduces the likelihood of misclassification, thus increasing the overall correctness of the test.
Problem 4: Stock price changes modeled as a binomial process
a. The probability that the stock decreases exactly $5 after 9 days (i.e., 4 up days and 5 down days) involves calculating the binomial probability with parameters n=9 and k=5 (down days): P = C(9,5) * (1/2)^9 ≈ 0.174. The number of ways to have exactly 5 down moves among 9 days is given by the binomial coefficient, and each sequence’s probability is assigned equally.
b. For the stock to have a net change of at most $3 (i.e., between -3 and +3 dollars), the number of up days should lie between 3 and 6. Summing probabilities across these ranges yields an approximate probability of 0.86, considering the symmetry and binomial distribution characteristics.
Problem 5: Expected value analysis of game options
Option A involves flipping 10 coins; winning occurs if the number of heads is 4, 5, or 6. The probability of this is calculated via binomial probabilities, summing P(X=4), P(X=5), P(X=6) with n=10 and p=0.5. The expected gain is then computed as the sum of each outcome's value weighted by its probability, resulting in a positive expected value, approximately $4.72.
Option B is the reverse, leading to an expected loss of similar magnitude. Option C involves no gamble, hence zero expected gain. Therefore, only Option A has a positive expected gain, making it the most favorable choice among the three.
In conclusion, the detailed analysis reveals that understanding the properties of probability distributions and binomial calculations is crucial for interpreting these scenarios accurately. Whether assessing outcomes of coin flips, stock movements, or strategic game options, the principles of probability provide essential insights into expected outcomes and decision-making strategies.
References
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- Ross, S. M. (2014). A First Course in Probability. Pearson.
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- Blitzstein, J., & Hwang, J. (2014). Introduction to Probability. CRC Press.
- Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
- Kendall, M., & Stuart, A. (1973). The Advanced Theory of Statistics. Charles Griffin & Co Ltd.
- Zwillinger, D. (2003). CRC Standard Probability and Statistics Tables and Formulae. CRC Press.
- McClave, J. T., & Sincich, T. (2018). Statistics. Pearson.
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