1. A single, standard number cube is tossed. What is the probability of getting a 4 or 5? (1 point)
1
one-fifth
one-third
start fraction 1 over 6 end fraction
start fraction 2 over 6 end fraction (or one-third) - because there are two out of six possible outcomes that result in a 4 or 5 (4 and 5), and there are six equally likely outcomes in total.
1. A single, standard number cube is tossed. What is the probability of getting a 4 or 5? (1 point)
1.1
2.one-fifth
3.one-third
4.start fraction 1 over 6 end fraction
The answer is 3. one-third. There are two possible outcomes that result in a 4 or 5 (4 and 5), out of a total of six possible outcomes (1, 2, 3, 4, 5, or 6). So, the probability of getting a 4 or 5 is 2/6, which simplifies to 1/3 or one-third.
2. A game involves spinning this spinner. What is the probability of the pointer landing on G?
A circle is divided equally into eight sections.· Two of the sections are labeled with an upper B.
· One of the sections is labeled with an upper G.
· One of the sections is labeled with an upper Y.
· Four of the sections are labeled with an upper R.
· An arrow originating from the center of the circle is pointing at one of the sections with an upper R.
(1 point)
one-eighth
start fraction 5 over 8 end fraction
one-half
start fraction 3 over 8 end fraction
The probability of the pointer landing on G is 1/8 or one-eighth. There is only one section on the spinner that is labeled with an upper G, and there are a total of eight equally likely sections on the spinner. Therefore, the probability of the pointer landing on G is 1/8.
3. The table shows the color preferences of 50 shoppers at the mall. What is the probability that one shopper, selected at random from the 50 surveyed, preferred the color red or blue?
Color Preferences
Color Number of Shoppers
Red 7
Yellow 3
Blue 13
Green 15
Orange 12
(1 point)
start fraction 7 over 50 end fraction
two-fifths
one-half
two-thirds
The probability that one shopper, selected at random from the 50 surveyed, preferred the color red or blue is:
P(red or blue) = P(red) + P(blue)
P(red) = 7/50 (from the table)
P(blue) = 13/50 (from the table)
P(red or blue) = 7/50 + 13/50
P(red or blue) = 20/50
P(red or blue) = 2/5
P(red or blue) = 0.4
Therefore, the probability that one shopper, selected at random from the 50 surveyed, preferred the color red or blue is two-fifths or 0.4.
4. A standard number cube with the numbers 1 through 6 is rolled. Find the probability of rolling a number greater than 5. (1 point)
start fraction 1 over 6 end fraction
one-third
one-fourth
two-thirds
The probability of rolling a number greater than 5 on a standard number cube is:
1/6
There is only one outcome (rolling a 6) out of six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6) that results in a number greater than 5. Therefore, the probability of rolling a number greater than 5 is 1/6.
5. A number cube is rolled 360 times, and the results are recorded as follows: 61 ones, 26 twos, 36 threes, 76 fours, 73 fives, and 88 sixes. What is the experimental probability of rolling a 2 or a 3? (1 point)
0.07
0.17
0.26
0.83
The experimental probability of rolling a 2 or a 3 is:
Number of times rolling a 2 or 3 / Total number of rolls
Number of times rolling a 2 or 3 = Number of twos + Number of threes = 26 + 36 = 62
Total number of rolls = 360
Experimental probability of rolling a 2 or a 3 = 62/360
Experimental probability of rolling a 2 or a 3 = 0.17 (rounded to two decimal places)
Therefore, the experimental probability of rolling a 2 or a 3 is 0.17. Answer: 2. 0.17.
6. From a barrel of colored marbles, you randomly select 5 blue, 2 yellow, 3 red, 4 green, and 2 purple marbles. Find the experimental probability of randomly selecting a marble that is not yellow. (1 point)
nine over sixteen
seven over eight
eight over nine
three over four
The total number of marbles is:
5 blue + 2 yellow + 3 red + 4 green + 2 purple = 16 marbles
The number of marbles that are not yellow is:
5 blue + 3 red + 4 green + 2 purple = 14 marbles
The experimental probability of randomly selecting a marble that is not yellow is:
Number of marbles that are not yellow / Total number of marbles
Experimental probability of selecting a marble that is not yellow = 14/16
Experimental probability of selecting a marble that is not yellow = 7/8
Therefore, the experimental probability of randomly selecting a marble that is not yellow is 7/8. Answer: 2. seven over eight.
7. The probability of winning a game is 25%. How many times should you expect to win if you play 20 times? (1 point)
7 times
10 times
2 times
5 times
If the probability of winning a game is 25%, the probability of losing the game is:
100% - 25% = 75%
If you play the game 20 times, the expected number of times you would win is:
Expected number of wins = Number of trials × Probability of winning
Expected number of wins = 20 × 25%
Expected number of wins = 5
Therefore, if you play the game 20 times, you can expect to win about 5 times. Answer: 4. 5 times.
8. A survey showed that 56% of car owners prefer four-door cars, 31% prefer two-door cars, and 13% have no preference. You ask 300 people. How many do you think will prefer four-door cars? (1 point)
218 people
168 people
233 people
68 people
If 56% of car owners prefer four-door cars and you are surveying 300 people, you can estimate the number of people who prefer four-door cars as:
Number of people who prefer four-door cars = Total number of people × Percentage who prefer four-door cars
Number of people who prefer four-door cars = 300 × 56%
Number of people who prefer four-door cars = 168
Therefore, based on the survey results, you can expect that about 168 people out of 300 will prefer four-door cars. Answer: 2. 168 people.
9. Mary works in a factory that produces 1,000 telephones each day. When 30 telephones were sampled, it was found that 9 were defective. Estimate how many telephones are defective each day. (1 point)
314 telephones
284 telephones
300 telephones
310 telephones
If 9 out of 30 telephones sampled were defective, this can be used to estimate the proportion of defective telephones in the entire daily production:
Proportion of defective telephones = Number of defective telephones sampled / Total number of telephones sampled
Proportion of defective telephones = 9/30
To estimate how many telephones are defective each day, this proportion can be applied to the total daily production of 1,000 telephones:
Number of defective telephones = Proportion of defective telephones × Total number of telephones produced
Number of defective telephones = (9/30) × 1,000
Number of defective telephones = 900/3
Number of defective telephones = 300
Therefore, based on the sample of 30 telephones, it can be estimated that about 300 telephones are defective each day. Answer: 3. 300 telephones.
10. The spinner is divided into equal parts. Use a tree diagram to find the probability that the spinner will land on a consonant both times if the spinner is spun twice.
A circle is divided equally into three sections.· One of the sections is labeled with an upper E.
· One of the sections is labeled with an upper U.
· One of the sections is labeled with an upper N.
· An arrow originating from the center of the circle is pointing at the section labeled with upper U. (1 point)
one-ninth
one-third
start fraction 5 over 9 end fraction
three-fourths
Here is the tree diagram for spinning the spinner twice:
```
U E N
/ \ / \ / \
C V C V C V
/ \ / \ / \
C V C V C V
```
Where U represents spinning a vowel on the first spin, C represents spinning a consonant on the second spin, V represents spinning a vowel on the second spin, E represents spinning an E on the first spin, and N represents spinning an N on the first spin.
The probability of spinning a consonant on the first spin is 2/3, since there are two consonants out of three possible outcomes. The probability of spinning a consonant on the second spin, given that a consonant was spun on the first spin, is 2/2, since all of the remaining outcomes are consonants. Therefore, the probability of spinning a consonant both times is:
P(consonant, consonant) = P(consonant on first spin) × P(consonant on second spin | consonant on first spin)
P(consonant, consonant) = (2/3) × (2/2)
P(consonant, consonant) = 2/3
Therefore, the probability of the spinner landing on a consonant both times if the spinner is spun twice is two-thirds or 0.67 (rounded to two decimal places). Answer: 2. one-third.
11. An ice cream shop offers 5 different flavors of ice cream and 12 different toppings. How many choices are possible for a single serving of ice cream with one topping? (1 point)
18 choices
22 choices
46 choices
60 choices
There are 5 different flavors of ice cream and 12 different toppings. To find the total number of choices for a single serving of ice cream with one topping, we can use the multiplication principle:
Total choices = Number of choices for ice cream × Number of choices for topping
Number of choices for ice cream = 5 (since there are 5 different flavors of ice cream)
Number of choices for topping = 12 (since there are 12 different toppings)
Total choices = 5 × 12
Total choices = 60
Therefore, there are 60 choices possible for a single serving of ice cream with one topping. Answer: 4. 60 choices.
12. Suzy likes to mix and match her 3 necklaces, 2 bracelets, and 6 hats. The colors are listed in the table. On Monday, she randomly picks a bracelet, a necklace, and a hat. What is the probability of Suzy choosing a red bracelet and silver hat?
Necklace Bracelet Hat
Red Red Silver
Green Black Yellow
Gold Green
Gold
Black
White
(1 point)
one-half
one-fourth
start fraction 1 over 6 end fraction
start fraction 1 over 12 end fraction
The probability of Suzy choosing a red bracelet and a silver hat can be found by dividing the number of ways she can choose a red bracelet and a silver hat by the total number of possible outcomes:
Total number of outcomes = 3 necklaces × 2 bracelets × 6 hats = 36
Number of ways to choose a red bracelet and a silver hat = 1 (there is only one red bracelet and one silver hat)
Therefore, the probability of Suzy choosing a red bracelet and a silver hat is:
P(red bracelet and silver hat) = Number of ways to choose a red bracelet and a silver hat / Total number of outcomes
P(red bracelet and silver hat) = 1 / 36
Therefore, the probability of Suzy choosing a red bracelet and a silver hat is 1/36. Answer: 4. start fraction 1 over 12 end fraction.
13. Which events are not independent? (1 point)
landing on heads after tossing a coin and rolling a 3 on a 6-sided number cube.
choosing a marble from a jar and landing on tails after tossing a coin
choosing a 5 from a deck of cards, replacing it, and then choosing an ace as the second card
choosing a card at random from a deck, without replacing it, and choosing another card at random
The events that are not independent are:
- Choosing a 5 from a deck of cards, replacing it, and then choosing an ace as the second card. These events are not independent because the outcome of the first selection affects the probability of selecting an ace on the second draw.
- Choosing a card at random from a deck, without replacing it, and choosing another card at random. These events are not independent because the outcome of the first selection affects the probability of selecting a certain card on the second draw, since there is one less card in the deck after the first selection.
The other two events mentioned are independent, since the outcome of one event does not affect the probability of the other event.
14. A coin is tossed and a number cube is rolled. What is the probability that the coin shows heads and the number cube shows 5? (1 point)
start fraction 1 over 6 end fraction
start fraction 1 over 12 end fraction
start fraction 2 over 3 end fraction
one-fourth
The probability of the coin showing heads is 1/2, and the probability of the number cube showing 5 is 1/6. To find the probability of both events occurring, we multiply these probabilities:
P(coin shows heads and number cube shows 5) = P(coin shows heads) × P(number cube shows 5)
P(coin shows heads and number cube shows 5) = (1/2) × (1/6)
P(coin shows heads and number cube shows 5) = 1/12
Therefore, the probability that the coin shows heads and the number cube shows 5 is 1/12. Answer: 2. start fraction 1 over 12 end fraction.
13. Which one of these events are not independent? (1 point)
landing on heads after tossing a coin and rolling a 3 on a 6-sided number cube.
choosing a marble from a jar and landing on tails after tossing a coin
choosing a 5 from a deck of cards, replacing it, and then choosing an ace as the second card
choosing a card at random from a deck, without replacing it, and choosing another card at random
The events that are not independent are:
- Choosing a 5 from a deck of cards, replacing it, and then choosing an ace as the second card. These events are not independent because the outcome of the first selection affects the probability of selecting an ace on the second draw.
- Choosing a card at random from a deck, without replacing it, and choosing another card at random. These events are not independent because the outcome of the first selection affects the probability of selecting a certain card on the second draw, since there is one less card in the deck after the first selection.
The other two events mentioned are independent, since the outcome of one event does not affect the probability of the other event.
but theres only one answer
My apologies, you are correct. The answer to question 13 is "Choosing a 5 from a deck of cards, replacing it, and then choosing an ace as the second card". These events are not independent because the outcome of the first selection affects the probability of selecting an ace on the second draw.
15. On Friday nights the local diner serves three main entrées, each with a choice of vegetable. The entrées are beef, chicken, and fish. The vegetables are spinach, broccoli, and carrots. How many possible dinners (entrée and vegetable combinations) does the diner serve? List them. (3 points)
There are three choices for entrées and three choices of vegetables. To find all the possible dinner combinations, we multiply the number of entrée choices by the number of vegetable choices:
3 entrées × 3 vegetables = 9 possible dinner combinations
Here are the possible dinner combinations:
1. Beef with spinach
2. Beef with broccoli
3. Beef with carrots
4. Chicken with spinach
5. Chicken with broccoli
6. Chicken with carrots
7. Fish with spinach
8. Fish with broccoli
9. Fish with carrots
Therefore, the diner serves 9 possible dinner combinations.