14 Sample Exam Problems
Exercise 14.1 (Problem 1 (5 points)) A box contains 3 red balls and 2 green balls. Two balls are drawn at random with replacement. What is the probability mass function of a variable that counts the number of red balls drawn?
The probability mass function of a discrete random variable is a function that gives the probability that the random variable takes each of its possible values. In this case, the random variable is the number of red balls drawn. The possible values of the random variable are 0, 1, and 2.
The probability to select a red ball is 3/(3+2) = 3/5 and the probability to select a green ball is 2/(3+2) = 2/5. Because of the replacement, the probability of selecting a red ball is the same in each draw and the draws are independent. Therefore, the probability mass function of the random variable is given by the binomial distribution:
P(X = k) = \binom{2}{k} \left(\frac{3}{5}\right)^k \left(\frac{2}{5}\right)^{2-k}, \quad k = 0, 1, 2.
where k is the number of red balls drawn (successes).
\begin{align*} P(X = 0) & = \binom{2}{0} \left(\frac{3}{5}\right)^0 \left(\frac{2}{5}\right)^{2} = 1 \times 1 \times \frac{4}{25} = \frac{4}{25}, \\ P(X = 1) & = \binom{2}{1} \left(\frac{3}{5}\right)^1 \left(\frac{2}{5}\right)^{1} = 2 \times \frac{3}{5} \times \frac{2}{5} = \frac{12}{25}, \\ P(X = 2) & = \binom{2}{2} \left(\frac{3}{5}\right)^2 \left(\frac{2}{5}\right)^{0} = 1 \times \frac{9}{25} \times 1 = \frac{9}{25}. \end{align*}
Exercise 14.2 (Problem 2 (5 points)) You are conducting a financial audit of a company. You have a list of 50 transactions, and you take a random sample (without replacement) of 6 transactions. What is the probability that you find at least one fraudulent transaction in your sample given that 10 of the transactions in the list are fraudulent?
The probability of finding at least one fraudulent transaction in the sample is equal to 1 minus the probability of finding no fraudulent transactions in the sample. The probability of finding no fraudulent transactions in the sample is the probability of selecting 6 transactions from the 40 non-fraudulent transactions. Let A_i be the event that the i-th transaction is fraudulent. The probability of finding no fraudulent transactions in the sample is:
\begin{align*} P(\text{no fraudulent transactions}) & = P(\bar{A}_1 \cap \bar{A}_2 \cap \ldots \cap \bar{A}_6) \\ & P(\bar{A}_1) \times P(\bar{A}_2 | \bar{A}_1) \times \ldots \times P(\bar{A}_6 | \bar{A}_1 \cap \bar{A}_2 \cap \ldots \cap \bar{A}_5) \\ & = \frac{40}{50} \times \frac{39}{49} \times \ldots \times \frac{35}{45}. \end{align*}
The probability of finding at least one fraudulent transaction in the sample is:
\begin{align*} P(\text{at least one fraudulent transaction}) & = 1 - P(\text{no fraudulent transactions}) \\ & = 1 - \frac{40}{50} \times \frac{39}{49} \times \ldots \times \frac{35}{45}. \end{align*}
Exercise 14.3 (Problem 3 (2 points)) The events A and B are independent. The probability of event A is 0.3 and the probability of event B is 0.4. What is the probability of the union of events A and B?
The probability of the union of two events is given by:
P(A \cup B) = P(A) + P(B) - P(A \cap B).
Because the events A and B are independent, the probability of the intersection of the two events is given by:
P(A \cap B) = P(A) \times P(B).
Therefore, the probability of the union of events A and B is:
P(A \cup B) = 0.3 + 0.4 - 0.3 \times 0.4 = 0.58.
Exercise 14.4 (Problem 4 (3 points)) 60 percent of the clients of a bank have a checking account, 40 percent have a savings account, and 20 percent have both. What is the probability that a randomly selected client has a checking account or a savings account? What is the probability that a randomly selected client has neither a checking account nor a savings account?
Let A be the event that a client has a checking account and B be the event that a client has a savings account. The probability of the union of events A and B is given by:
P(A \cup B) = P(A) + P(B) - P(A \cap B).
The probability of the intersection of the two events is given by:
P(A \cap B) = 0.2.
Therefore, the probability that a randomly selected client has a checking account or a savings account is:
P(A \cup B) = 0.6 + 0.4 - 0.2 = 0.8.
The probability that a randomly selected client has neither a checking account nor a savings account is given by:
P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) = 1 - 0.8 = 0.2.
Exercise 14.5 (Problem 5 (5 points)) The Chebychev inequality states that for any random variable X with mean \mu and variance \sigma^2, and for any k > 0, the probability that X takes a value within k standard deviations of the mean is at least 1 - 1/k^2.
P(|X - \mu| \leq k\sigma) \geq 1 - \frac{1}{k^2}
Use the Chebychev inequality to find a lower bound on the probability that a random variable with mean 10 and variance 4 takes a value between 4 and 16. Why is the Chebychev inequality only useful for k > 1?
The Chebychev inequality states that the probability that a random variable takes a value within k standard deviations of the mean is at least 1 - 1/k^2. In this case, the mean of the random variable is \mu = 10 and the standard deviation is \sigma = \sqrt{4} = 2. The probability that the random variable takes a value between 4 and 16 is:
P(4 \leq X \leq 16) = P(|X - 10| \leq 3 \times 2) = P(|X - 10| \leq 6) \geq 1 - \frac{1}{3^2} = \frac{8}{9}.
The Chebychev inequality is only useful for k > 1 because the probability that a random variable takes a value within k standard deviations of the mean is always at least 0. Therefore, the Chebychev inequality is not useful for k = 1 because the probability that a random variable takes a value within 1 standard deviation of the mean is always at least 0.
Exercise 14.6 (Problem 6 (5 points)) In a game of “Who wants to be a millionaire?”, a contestant is asked a series of multiple-choice questions. Each question has four possible answers, and only one of them is correct. Assume that ten thousand contestants are playing the game, and each contestant randomly guesses the answer to each question, i.e. they pick an answer with equal probability among the four choices. How many of the 100 contestants would expect to win the grand prize of one million dollars if the game has 5 questions?
The probability that a contestant wins the grand prize is the probability that the contestant answers all questions correctly. The probability that a contestant answers a question correctly simply by guessing is 1/4. Because the contestants are guessing the answers, the answers to the questions are independent. Therefore, the probability that a contestant answers all questions correctly is:
\left(\frac{1}{4}\right)^5 = \frac{1}{1024}.
The expected number of contestants that would win the grand prize is the number of contestants times the probability that a contestant wins the grand prize:
10000 \times \frac{1}{1024} \approx 9.77.
Therefore, we would expect that about 10 of the 10,000 contestants would win the grand prize.
Exercise 14.7 (Problem 7 (5 points)) You conduct a survey using a random sample of 1000 persons with the goal to estiamate the proportion of people in a city who are in favor of a new policy. You find that 60 persons in the sample are in favor of the policy. Use the central limit theorem to find an approximate 95% confidence interval for the proportion of people in the city who are in favor of the policy.
The central limit theorem states that the distribution of the sample mean of a random sample from a population approaches a normal distribution as the sample size increases. The sample proportion of people in favor of the policy is \hat{p} = 60/1000 = 0.06. The standard error of the sample proportion is given by:
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.06 \times 0.94}{1000}} \approx 0.0077.
The 95% confidence interval for the population proportion is given by:
\hat{p} \pm 1.96 \times SE = 0.06 \pm 1.96 \times 0.0077 = (0.044, 0.076).
Therefore, we are 95% confident that the proportion of people in the city who are in favor of the policy is between 0.044 and 0.076.