Name:

  1. Suppose you flip a fair coin 4 times. Answer the following questions based on this information.
  1. Write out each element in the sample space.
  1. HHHH
  2. HHHT
  3. HHTH
  4. HTHH
  5. THHH
  6. HHTT
  7. HTHT
  8. TTHH
  9. THTH
  10. HTTH
  11. THHT
  12. TTTH
  13. TTHT
  14. THTT
  15. HTTT
  16. TTTT
  1. How many ways can exactly two tails appear in the experiment?

4 Choose 2 = 6

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

2^2 = 4

  1. Leo’s Pizza has 12 distinct toppings to choose from.
  1. How many different three topping pizzas are possible?

Combinations 12 Choose 3 = 220

  1. How many different four topping pizzas are possible?

Combinations 12 Choose 4 = 495

  1. Most online accounts now require seven characters to generate a strong password. To simplify, imagine you can only draw from the 26 letters in the alphabet and there is no difference between capital or lower case letters. How many unique seven character passwords are possible?

26^8 = 208827064576

  1. How many seven-digit unique phone numbers are possible for one area code?

10^7 = 10,000,000

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

Permutations (Order Matters) 12 Permut 12 = 479001600

  1. Consider an 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 six tosses and the number of tails in another column that occur in the six-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 what we did in class). In a fourth column compute the probabilities of each event. Use the table to answer the following questions:
x <- seq(from = 0, to = 8, by = 1)
n <- length(x)

N <- 2^8

count_x <- double(n)
prop <- double(n)
for(i in 1:n) {
  count_x[i] <- dim(combn(8,x[i]))[2]
  prop[i] <- count_x[i]/N
}

dat <- data.frame(x,count_x,prop)

dat
##   x count_x    prop
## 1 0       1 0.00391
## 2 1       8 0.03125
## 3 2      28 0.10938
## 4 3      56 0.21875
## 5 4      70 0.27344
## 6 5      56 0.21875
## 7 6      28 0.10938
## 8 7       8 0.03125
## 9 8       1 0.00391
ans_c <- 1-prop[1]

ans_e <- sum(prop[1:5])
  1. What are the total number of elements in the sample space?

256

b.Create the table in excel.

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

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

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

\(P(T = 0) = 0.0039\)

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

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

  1. Recall that the probability of getting exactly one head out of two tosses is, , and the probability of getting two heads out of four tosses is . Determine the probability of getting exactly four heads out of eight. In each case we are looking at the probability for 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.

As you increase the number of flips the total number of outcomes is increasing. While the highest probability is exactly half heads that probability is declining as the number of flips increases.

  1. A survey of magazine subscribers showed that 45.8% rented a car during the past 12 months for business reasons, 54% rented a car during the past 12 months for personal reasons, and 30% rented a car during the past 12 months for both business and personal reasons.
  1. What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons?
  2. What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?

A = Rented car for business reasons P(A) = .458 B = Rented car for personal reasons P(B) = 0.54

P(A and B) = 0.3

  1. Prob rented a car for business or personal reasons: P(A or B) = P(A) + P(B) - P(A and B) P(A or B) = 0.458 + 0.54 - 0.4 P(A or B) = 0.698

  2. Complement - everyone who didn’t rent a car for business or personal reasons 1 - P(A or B) = 0.302