Unit 3.1 Stats Quiz

An American roulette wheel has 38 slots with numbers 1 through 36, 0, and 00, as shown in the figure. Of the numbered slots, 18 are red, 18 are black, and 2—the 0 and 00—are green. When the wheel is spun, a metal ball is dropped onto the middle of the wheel. If the wheel is balanced, the ball is equally likely to settle in any of the numbered slots. Imagine spinning a fair wheel once. Define events B: ball lands in a black slot, and E: ball lands in an even-numbered slot. (Treat 0 and 00 as even numbers.) Make a two-way table that displays the sample space in terms of events B and E. Find P(B) and P(E). Describe the event "B and E" in words. Then find the probability of this event. Explain why P(B or E) ≠ P(B) + P(E). Then use the general addition rule to compute P(B or E).

P(B) = 18/38 = 0.474; P(E)= 20/38 = 0.526. The event "B and E" would be that ball lands in a spot that is black and even. P (B and E) = 10/38 = 0.263. The probability of the event "B or E" means the probability of landing in a spot that is either black, even, or both. If we add the probabilities of landing in a black spot and landing in an even spot, the spots that are black and even will be double counted because events B and E are not mutually exclusive. P(B and E) = 18/38 + 20/38 - 10/ 38 = 28/38 = 0.737

The Pew Research Center asked a random sample of 2024 adult cell-phone owners from the United States their age and which type of cell phone they own: iPhone, Android, or other (including non-smartphones). The two-way table summarizes the data. Suppose we select one of the survey respondents at random. What's the probability that: The person is not age 18 to 34 and does not own an iPhone? The person is age 18 to 34 or owns an iPhone?

P(not age 18 to 34 and not own an iPhone) = 189+100+277+643/2024 = 1209/2024 = 0.403

Suppose A and B are two events such that P(A) = 0.3, P(B) = 0.4, and P(A ∪ B) = 0.58. Find P(A ∩ B).

The general addition rule states that P(A ∪ B) = P(A) + P(B) − P(A ∩ B). We know that P(A) = 0.3, P(B) = 0.4, and P(A ∪ B) = 0.58. Therefore, 0.58 = 0.3 + 0.4 − P(A ∩ B). Solving for P(A ∩ B) gives P(A ∩ B) = 0.12.

In one large city, 40% of all households own a dog, 32% own a cat, and 18% own both. Suppose we randomly select a household. What's the probability that the household owns a dog or a cat?

P(own a dog or a cat) = P (Own a cat) - P(own a dog and a cat) = 0.40 + 0.32 - 0.18= 0.54

A four-sided die is a pyramid whose four faces are labeled with the numbers 1, 2, 3, and 4 (see image). Imagine rolling two fair, four-sided dice and recording the number that is showing at the base of each pyramid. For instance, you would record a 4 if the die landed as shown in the figure. A. Give a probability model for this chance process. B. Define event A as getting a sum of 5. Find P(A).

A. The following table shows the possible outcomes in the sample space. Each of the 16 outcomes will be equally likely and have probability 1/16. B. P(A) = 0.25. There are four ways to get a sum of 5 from these two dice: (1, 4), (2, 3), (3, 2), (4, 1). The probability of getting a sum of 5 is P(A)= 4/16= 0.25.

Choose a young adult (aged 25 to 29) at random. The probability is 0.13 that the person chosen did not complete high school, 0.29 that the person has a high school diploma but no further education, and 0.30 that the person has at least a bachelor's degree. What must be the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor's degree? Why? Find the probability that the young adult completed high school. Which probability rule did you use to find the answer? Find the probability that the young adult has further education beyond high school. Which probability rule did you use to find the answer?

The given probabilities have a sum of 0.72 and the sum of all probabilities should be 1. Thus, the probability that a randomly chosen young adult has some education beyond high school but does not have a bachelor's degree is 1 − 0.72 = 0.28. There is a 0.28 probability that a young adult has some education beyond high school but does not have a bachelor's degree. (b) Using the complement rule, P(at least a high school education) = 1 − P(has not finished high school) = 1 − 0.13 = 0.87. There is a 0.87 probability that a young adult has completed high school. (c) P(young adult has further education beyond high school) = P(young adult has some education beyond high school but does not have a bachelor's degree) + P(young adult has at least a bachelor's degree) = 0.28 + 0.30 = 0.58. The probability rule used is the addition rule for mutually exclusive events

In government data, a household consists of all occupants of a dwelling unit. Choose an American household at random and count the number of people it contains. Here is the assignment of probabilities for the outcome. The probability of finding 3 people in a household is the same as the probability of finding 4 people. What probability should replace "?" in the table? Why? Find the probability that the chosen household contains more than 2 people.

The probabilities of all the possible outcomes must add to 1. The given probabilities add up to 0.25 + 0.32 + 0.07 + 0.03 + 0.01 = 0.68. This leaves a probability of 1 − 0.68 = 0.32 for P(3 or 4). Because the probability of finding 3 people in a household is the same as the probability of finding 4 people (and they are mutually exclusive), each probability must be 0.32/2 = 0.16. (b) P(more than 2 people) = 1 − P(1 or 2 people) = 1 − (0.25 + 0.32) = 0.43. This could also be found using the addition rule for mutually exclusive events. P(more than 2 people) = P(3 or 4 or 5 or 6 or 7+) = 0.16 + 0.16 + 0.07 + 0.03 +0.01 = 0.43.

Mr. Starnes and his wife have 6 grandchildren: Connor, Declan, Lucas, Piper, Sedona, and Zayne. They have 2 extra tickets to a holiday show, and will randomly select which 2 grandkids get to see the show with them. A. Give a probability model for this chance process. B. Find the probability that at least one of the two girls (Piper and Sedona) get to go to the show.

The sample space is: Connor/Declan, Connor/Lucas, Connor/Piper, Connor/Sedona, Connor/Zayne, Declan/Lucas, Declan/Piper, Declan/Sedona, Declan/Zayne, Lucas/Piper, Lucas/Sedona, Lucas/Zayne, Piper/Sedona, Piper/Zayne, and Sedona/Zayne. Each of these 15 outcomes will be equally likely and will have probability 1/15. (b) There are 9 outcomes in which Piper or Sedona (or both) get to go to the show: Connor/Piper, Connor/Sedona, Declan/Piper, Declan/Sedona, Lucas/Piper, Lucas/Sedona, Piper/Sedona, Piper/Zayne, Sedona/Zayne. Define Event A as Piper or Sedona (or both) get to go to the show. Then P(A) = 9/15 = 0.60.

Ms. Tyson keeps a Mystery Box in her classroom. If a student meets expectations for behavior, she or he is allowed to draw a slip of paper without looking. The slips are all of equal size, are well mixed, and have the name of a prize written on them. One of the "prizes"—extra homework—isn't very desirable! Here is the probability model for the prizes a student can win: Explain why this is a valid probability model. Find the probability that a student does not win extra homework. What's the probability that a student wins candy or a homework pass?

This is a valid probability model because each probability is between 0 and 1 and the probabilities sum to 1. (b) P(student won't win extra homework) = 1 − 0.05 = 0.95. There is a 0.95 probability that a student will not win extra homework. (c) P(candy or homework pass) = 0.25 + 0.15 = 0.40. There is a 0.40 probability that a student wins candy or a homework pass.