EC245 - Assignment 3 - Key

  1. Suppose you flip a fair coin 3 times. Use this to answer the following questions:
  1. What is the sample space for this experiment? Write out each element. \[S = \{HHH,HHT,HTH,THH,HTT,TTH,THT,TTT\}\]
  2. How many ways can exactly two tails appear in the experiment?

There are three ways exactly two tails can appear in the experiment. You can count them from the set or use combinations.

  1. How many different outcomes are possible if the first two tosses have heads?

There are two possible outcomes given the first two tosses are heads: {HHH} or {HHT}

  1. Leo’s Pizza has 12 distinct toppings to choose from.

  1. How many different three topping pizzas are possible? Order doesn’t matter: COMBIN(12,3) = 220

  2. How many different four topping pizzas are possible? Order doesn’t matter: COMBIN(12,4) = 495

  1. How many unique six letter passwords are possible drawing only from the English alphabet lowercase letters?

Basic Rule of Counting: \(n = 26^{6} = 308915776\)

  1. How many seven digit unique phone numbers are possible?

Basic Rule of Counting: \(n = 10^{7} = 10000000\)

  1. How many ways is it possible to assign seats in a 12 student class?

Permutations: \(n = permut(12,12) = 479001600\)

  1. How many ways is it possible to select the first and second ranked students in a class of 10?

Permutations: \(n = permut(10,2) = 90\)

  1. Consider the experiment of flipping a fair coin 8 times. First, determine the total number of elements or sample points in the sample space. Second, using excel, create a table of two columns showing the number of heads in one column that occur in the 8 tosses and the number of tails in another column that occur in the 8 coin flips. In a third column compute the total number of combinations that occur for each number of heads and tails in the table (similar to the Week 3 Video 2). In a fourth column compute the probabilities of each event. Use the table to answer the following questions:

Total number of outcomes: \(n = 2^{8} = 256\)

  1. What is the probability of obtaining at least one tails in the 8 flips?

\(P(x \geq 1) = 0.996\)

  1. What is the probability of obtaining no tails in the 8 flips?

\(P(x = 0) = 0.004\)

  1. What is the probability of obtaining at most four heads in the 8 flips?

\(P(x \leq 4) = 0.637\)

  1. Recall that the probability of getting exactly one head out of two tosses is, p = 0.5 , and the probability of getting two heads out of four tosses is p = 0.375. Determine the probability of getting exactly four heads out of eight. In each case we are looking at the probability of getting exactly half heads in the experiment. Comment on what is happening to the probability of getting exactly half heads as we increase the number of flips. Explain why this trend is occurring.

\(P(x = 4) = 0.273\)

The probability of getting exactly half heads is declining as we increase the number of tosses. This is a result of the number of possible outcomes increasing as we increase the tosses. As there are more possible outcomes the likelihood of getting exactly half is decreasing. It should be noted that there is still a high likelihood of getting a number of heads close to half but not exactly half.

  1. In the past month 45.8% of rental car customers rented a car for business reasons, 54% rented a car for personal reasons, and 30% rented a car for business and personal reasons.
  1. What is the probability that a customer rented a car for business or personal reasons?

\(P(A \cup B) = 0.698\)

  1. What is the probability that a customer did not rent a car for either business or personal reasons?

\(1-P(A \cup B) = 0.302\)

  1. Suppose that we have two events, A and B, with P (A) = .50, P (B) = .60, and P (A|B) = .30.
  1. Find \(P(A|B) = \frac{0.3}{0.6} = 0.5\)
  2. Find \(P(B|A) = \frac{0.3}{0.5}= 0.6\)
  1. Assume that we have two events, A and B, that are mutually exclusive. Assume that we know P (A) = .30 and P (B) = .40.
  1. What is \(P(A \cap B) = 0\)
  2. What is \(P(A|B) = 0\)
  1. A recent study looked to examine health insurance coverage by age. The following contingency table contains counts for two categorical variables: Age (18-34 or 35 and older) and Health Insurance Coverage (Yes or No). Use this information to answer the following questions.

Health Insurance

Yes No Total
18 to 34 0.375 0.085 0.46
35 and Older 0.475 0.065 0.54
Total 0.85 0.15 1

  1. Develop a joint probability table for these data and use the table to answer the remaining questions.

  2. What do marginal probabilities tell you about the age of the U.S. population?

The population is relatively older as 54% are characterized as 35 and older and only 46% are characterized as 18 to 34.

  1. What is the probability that a randomly selected individual does not have health insurance coverage?

The probability that a randomly selected individual does not have health insurance coverage is 15%.

  1. If the individual is between the ages of 18 and 34, what is the probability that the individual does not have health insurance coverage?

Conditional Probability: \(P(No Ins|18-34) = \frac{0.085}{0.46} = 0.185\)

  1. If the individual is age 35 or older, what is the probability that the individual does not have health insurance coverage?

Conditional Probability: \(P(No Ins|\geq 35) = \frac{0.065}{0.54} = 0.12\)

  1. If the individual does not have health insurance, what is the probability that the individual is in the 18-34 group?

Conditional Probability: \(P(18-34|NoINs) =\frac{0.085}{0.15}= 0.57\)

  1. What does the probability information tell you about health insurance coverage in the United States?

Young people are not insured at the same rate as the older population.