CHAPTER 7 PROB TEST

In a game, a 6-sided die and spinner with 10 equal spaces numbered 1 through 10 will be spun. What is the probability for the die to be even and the spinner to be greater than 6 ? a) 1/5 b) 2/5 c) 1/2 d) 3/4

A. 1/5 The probability of an even number is 3/6 or 1/2. The probability for a spinner to be greater than 6 is 4/10 or 2/5. The probabilities are multiplied together. Therefore, (1/2)(2/5)=2/10=1/5.

A 20-sided fair die is rolled for a game, and the sides are numbered 1 through 20. What is the probability of rolling a multiple of 9 or an even number? a) 11/20 b) 1/10 c) 1/5 d) 1/2

A. 11/20 The probability of a multiple of 9 is 2/20 and rolling an even number is 10/20. There is 1 value that is a multiple of 9 and even (18). The probability is 2/20+10/20−1/20=12/20−1/20=11/20.

Suppose the probability of having a girl is 50%. If a study finds that the probability of being interested in sports if you are a girl is 60%, what is the probability of parents giving birth to a girl who is interested in sports? a) 30% b) 55% c) 80% d) 110%

A. 30% Use the general multiplication rule to find the answer. Since the probability of having a girl is 0.5, and the probability of the girl liking sports is 0.6, then 𝑃(being a girl and liking sports)=.5×.6

According to the United Nation (2016), global population was distributed as follows: 60% of the world's population lives in Asia, 16% in Africa, 10% in Europe, 8% in North America, 6% in South America, 1% in Oceania. What's the probability that a randomly selected person will live in the Americas? A. 6% b) 8% c) 14% d) 15%

A. 6% The probability is the sum of the Americas (North America and South America), which is 8%+6%.

There are 1,000 raffle tickets for a fundraiser. If a mother buys 25 tickets, and a father buys 20 tickets, what is the probability the family will win? a) 9/200 b) 9/100 c) 9/50 d) 9/40

A. 9/200 There are 45 tickets purchased. The probability is 45/1,000=9/200.

In asking the question "what is the probability Jake will hit a home run on Saturday?," "hitting a home run" would be called... a) the event b) the probability c) the sample space d) the complement e) disjoint

A. The Event The event is the outcome whose likelihood is being measured by the probability. Therefore, in this question "hitting a home run" would be the event.

Two events that cannot happen simultaneously in the same experiment are called disjoint. a) True b) False

A. True The correct term for two events that cannot both be true is disjoint.

For each day in New England, the probability that it is mostly cloudy is 0.10. Cindi is planning to go leaf watching, and her friend will join her with probability 0.70. Find the probability that when Cindi takes her trip, either Cindi's friend does not join her, or she does join her but it is mostly cloudy. Assume that her friend joining her is independent of the weather conditions. a) 0.1 b) 0.37 c) 0.77 d) 0.703

B. 0.37 The probability that the friend doesn't come is 0.3. The probability that she comes and it's cloudy is 0.7×0.1=0.07. The probability of one or the other of these events is 0.3+0.07=0.37.

There are 52 cards in a deck. The deck has 4 sets of cards numbered 2 through 10, as well as 4 sets of an ace, king, queen, and jack. What is the probability of an odd card first and jack, king, or queen card drawn second if the cards are not replaced? a) 20/663 b) 16/221 c) 4/17 d) 5/13

B. 16/221 There are 4 odd cards in each of the 4 suits (3,5,7,9), so the probability of an odd card is 16/52 or 4/13. The probability of a jack, king, or queen is 12/51 or 4/17. The probabilities are multiplied together since both events need to happen. Therefore, (4/13)(4/17)=16/221.

On any given day the probability that Zoren will drive past the local bakery is 1/3. If he has enough money, then he will stop to buy a tasty treat. Fortunately, the probability that he has enough money is 4/5. Assuming that the events in question are independent, find the probability that on a given day Zoren will buy a tasty treat from the local bakery. a) 1/15 b) 4/15 c) 11/15 d) 1

B. 4/15 This is an "and" question which uses the formula 𝑃(𝐴)×𝑃(𝐵).

A pair of six-sided dice are rolled. What is the theoretical probability the sum of the dice is greater than 8? a) 1/18 b) 5/18 c) 7/18 d) 11/18

B. 5/18 There are 36 outcomes when rolling a pair of die and 10 of those outcomes sum to greater than 8. The probability is 10/36 or 5/18.

Roger has two blue ties, one red tie, two blue coats, and two orange coats. What is the probability that he wears a blue tie or a blue coat? a) 2/3 b) 5/6 c) 1/2 d) 1/3

B. 5/6 𝑃(blue tie) is 2/3. 𝑃(blue coat)=1/2. The 𝑃(blue tie or blue coat)=2/3+1/2−1/3=5/6.

The numbers 1 through 20 are in a bag. What is the probability of drawing a 6 first, then an even number without replacement? a) 9/400 b) 9/380 c) 9/200 d) 9/190

B. 9/380 The probability of a 6 is 1/20. The probability of an even number is 9/19. The probabilities are multiplied together since both events need to happen. Therefore,(1/20)(9/19)=9/380.

A student has the probability of walking to school of 0.3, a probability of riding a bike of 0.5, and a probability of riding in a car of 0.2. The probability for a student being late is 0.1 when walking, 0.05 when riding a bike, and 0.01 when riding in a car. What is the probability of walking and being late or riding in a car and being late? a) 0.002 b) 0.03 c) 0.032 d) 0.05

C. 0.032 The probability is the sum of the product of each method and being late. This is 0.3(0.1)+0.2(0.01)=0.03+0.002=0.032.

There are 4 baseball games during the week. The probability of a rainout is 10% for each game. Assuming the weather each day is independent, what is the probability at least one game will not rainout? a) 0.0001 b) 0.656 c) 0.9999 d) 0.344

C. 0.9999 The probability of the complementary event (there are only rainouts) is 0.1*0.1*0.1*0.1=0.0001. So the probability that there is at least one without a rainout will be 1−0.0001=0.9999

A roulette wheel has 38 spaces. There are 2 even number green spaces, 10 odd red spaces, 8 red even spaces, 10 even black spaces, and 8 odd black spaces. What is the probability that the wheel will not land on a red number? a) 7/19 b) 9/19 c) 10/19 d) 12/19

C. 10/19 There are 20 spaces that are not red. Therefore, the probability is 20/38 or 10/19.

In a bag, there are 3 yellow, 4 green, and 2 blue marbles. What is the probability of choosing a green marble? a) 3/4 b) 4/7 c) 4/9 d) 2/9 e) 1/3

C. 4/9 There are 4 green marbles and 9 total marbles.

There are 2 suits of cards, spades and clubs, and 13 cards in each suit, numbered 2 through 10 inclusive as well as a jack, queen, king and ace. What is the probability that someone pulls a spade that is NOT a jack, queen, king or ace? a) 4/13 b) 10/26 c) 9/26 d) 4/26

C. 9/26 There are 26 cards in the deck and there are 9 non-face card spades. So the probability is 9/26.

At a restaurant for lunch, there are 4 sandwich choices, 6 side choices, and 8 drink choices. What is the size of the sample space if one choice is selected from each? a) 24 b) 32 c) 164 d) 192

D. 192 There are 4×6×8=192 choices in the sample

Angelique wants to go on a vacation, but she's anxious about whether or not she's going to get a year-end bonus at work. She knows that she is less likely to get a bonus if she doesn't meet her sales target. She has a 5.88% chance of missing her sales target, and 60% of associates who miss their targets do not receive a bonus. What is Angelique's probability of missing her sales target, and subsequently not receiving a bonus? a) 1.32% b) 1.98% c) 2.35% d) 3.5%

D. 3.5% The relevant information here is that she has a 5.88% chance of missing her sales target and a 60% chance if she does miss her target of also missing out on a bonus. The probability of missing both her sales target and not receiving a bonus is 0.0588×0.6=3.5%

A restaurant has a small, medium, and large pizza, 4 types of cheese, 5 types of sauces, 10 meat toppings, and 15 vegetable toppings. What is the size of the sample space for a 1 cheese, 1 meat topping, and 1 type of sauce pizza? a) 20 b) 60 c) 200 d) 600

D. 600 There are 3 sizes, 4 cheeses, 10 meat toppings and 5 sauces. Therefore, there are 3×4×5×10=600 pizzas in the sample space.

According to the United Nation (2016), global population was distributed as follows: 60% of the world's population lives in Asia, 16% in Africa, 9% in Europe, 8% in North America, 6% in South America, 1% in Oceania. What's the probability that a randomly selected person does NOT live in the Americas? a) 14% b) 30% c) 70% d) 86%

D. 86% The probability is the sum of both continents in the Americas, subtracted from 1, so 1−.08−.06=.86.

If the probability of event 𝐸 is 4/9, the probability of the complement of 𝐸 is 5/9. a) True b) False

True This statement is true. The probability of an event and its complement sums to 1, so if the probability of event 𝐸 is 4/9, the probability of the complement of 𝐸 is 5/9.

There are 18 red, 15 white, and 12 blue pieces of candy in a bag. What is the probability of selecting 2 blue, 1 red, and 1 white piece, replacing the candy each time? a) 8/3,375 b) 16/3,375 c) 32/3,375 d) 486/3,375

c) 32/3,375 There are 45 pieces of candy in the bag. The probability of one blue is 12/45 or 4/15. The probability of one red is 18/45 or 2/5. The probability of one white is 15/45 or 1/3. Since there are 2 blue, 1 red, and 1 white drawn, the probability is multiplied 4 times. Therefore, (4/15)(4/15)(2/5)(1/3)=32/3375.

Keri can get to work on time when the trains are running or when her car is working. Let 𝐴= the trains are running and 𝐵= her car is working. a) 𝑃(𝐴 and 𝐵)=𝑃(𝐴)×𝑃(𝐵|𝐴) b) 𝑃(𝐴 or 𝐵)=𝑃(𝐴)+𝑃(𝐵) c) 𝑃(𝐴 and 𝐵)=𝑃(𝐴)×𝑃(𝐵) d) 𝑃(𝐴 or 𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴 and 𝐵)

d) 𝑃(𝐴 or 𝐵)=𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴 and 𝐵) Keri can get to work on time when the trains are running or when her car is working, or both. Since these events are not disjoint, the correct answer is (d).