The letters for the word math are placed in a bag. Two letters are selected with replacement. Which statements are true about the tree diagram that could represent the possible outcomes?? Select all that apply.
A. There would be 4 tree bases.
B. Each base would have 3 tree branches.
C. The diagram would show a total of 12 possible outcomes.
D. There would be 4 combinations that have double letters.
A. There would be 4 tree bases.
B. Each base would have 3 tree branches.
C. The diagram would show a total of 16 possible outcomes.
The letters for the word math are placed in a bag. Two letters are selected with replacement. Which statements are true about the tree diagram that could represent the possible outcomes? Select all that apply. Only can chose 2
A. There would be 4 tree bases.
B. Each base would have 3 tree branches.
C. The diagram would show a total of 12 possible outcomes.
D. There would be 4 combinations that have double letters.
A. There would be 4 tree bases.
B. Each base would have 4 tree branches.
A spinner has three equal sections labeled 1, 2, and 3.
Which organized list shows the possible outcomes for spinning the spinner twice?
A.
1,1; 1,2; 1,3
2,1; 2,2; 2,3
3,1; 3,2; 3,3
B.
1,2; 1,3
2,1; 2,3
3,1; 3,2
C.
1,1; 1,2; 1,3
2,2; 2,3
3,3
D.
1,1; 1,2; 1,3
A.
1,1; 1,2; 1,3
2,1; 2,2; 2,3
3,1; 3,2; 3,3
A spinner has three equal sections labeled 1, 2, and 3.
What is the probability of getting two numbers that have a sum of 5 on the spinner?
A. two-ninths
B. one-third
C. one-sixth
D. four-ninths
4 / 4
3 of 4 Answered
B. one-third
The only way to get a sum of 5 is to spin a 2 and a 3, or to spin a 3 and a 2. The probability of spinning a 2 is 1/3 and the probability of spinning a 3 is 1/3. Since the spins are independent, the probability of getting both a 2 and a 3 is (1/3) × (1/3) = 1/9. But there are two possible ways to get a sum of 5, so the total probability is 2/9, which simplifies to one-third.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing a red marble, then a white marble if the marbles are replaced.
A. one-twelfth
B. five over thirty-six
C. five-sixths
D. five-twelfths
D. five-twelfths
The probability of choosing a red marble on the first draw is 5/12 because there are 5 red marbles out of a total of 12 marbles. Since the marble is replaced before the second draw, the probabilities are the same for the second draw. The probability of choosing a white marble on the second draw is 4/12, or 1/3, because there are 4 white marbles remaining out of a total of 12 marbles. To find the probability of both events happening, we multiply the probabilities: (5/12) × (1/3) = 5/36. But the question asks for the probability of first choosing a red marble, then choosing a white marble. Since the marble is replaced after the first draw, these events are independent, so we can multiply the probabilities again: (5/12) × (4/12) = 20/144, which simplifies to 5/36.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing 3 blue marbles in a row if the marbles are replaced.
A. two over eleven
B. one over two hundred twenty
C. Fraction 1 over 27 end fraction
D. one over sixty-four
B. one over two hundred twenty
The probability of choosing a blue marble on any given draw is 3/12, or 1/4, because there are 3 blue marbles out of a total of 12 marbles. Since the marbles are replaced between draws, the probabilities are the same for each draw. To find the probability of choosing 3 blue marbles in a row, we multiply the probabilities of each individual draw: (1/4) × (1/4) × (1/4) = 1/64. So the probability of choosing 3 blue marbles in a row with replacement is 1/64.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing a blue marble, then a red marble if the marbles are not replaced.
A. five over forty-four
B. fifteen over thirty-five
C. two-thirds
D. one over fifteen
B. fifteen over thirty-five
The probability of choosing a blue marble on the first draw is 3/12, or 1/4, because there are 3 blue marbles out of a total of 12 marbles. Since the marble is not replaced before the second draw, the probability of choosing a red marble on the second draw depends on the outcome of the first draw. After one blue marble has been removed, there are 11 marbles left in the bag, of which 5 are red. So the probability of choosing a red marble on the second draw, given that a blue marble was chosen on the first draw, is 5/11. To find the probability of both events happening, we multiply the probabilities: (1/4) × (5/11) = 5/44. But the question asks for the probability of first choosing a blue marble, then choosing a red marble. Since the marbles are not replaced after the first draw, these events are dependent, so we need to multiply by the probability of choosing a blue marble on the first draw again, which is 1/4: (1/4) × (5/11) × (1/4) = 5/176. Simplifying this fraction gives us 15/35, or three-fifths.
A bag contains 4 white, 3 blue, and 5 red marbles.
Find the probability of choosing 2 white marbles in a row if the marbles are not replaced.
A. The fraction states 1 over 11.
B. one-ninth
C. two-thirds
D. Start Fraction 1 over 16 End Fraction
D. Start Fraction 1 over 16 End Fraction
The probability of choosing a white marble on the first draw is 4/12, or 1/3, because there are 4 white marbles out of a total of 12 marbles. Since the marble is not replaced before the second draw, the probability of choosing a white marble on the second draw depends on the outcome of the first draw. After one white marble has been removed, there are 11 marbles left in the bag, of which 3 are white. So the probability of choosing a white marble on the second draw, given that a white marble was chosen on the first draw, is 3/11. To find the probability of both events happening, we multiply the probabilities: (1/3) × (3/11) = 3/33. But the question asks for the probability of first choosing a white marble, then choosing another white marble. Since the marbles are not replaced after the first draw, these events are dependent, so we need to multiply by the probability of choosing a white marble on the second draw again, which is 3/11: (1/3) × (3/11) × (1/3) = 1/33. Simplifying this fraction gives us 1/16.
One-half of the students at Greendale Middle School take art. Two-thirds of the students take Spanish. Deshawn uses a coin and a standard number cube as a simulation of the students at Greendale Middle School.
Deshawn lets “T” represent a student taking art and “H” represent a student not taking art. He lets the numbers 1, 2, 3, and 4 on the number cube represent a student taking Spanish, and the numbers 5 and 6 represent a student not taking Spanish. The table shows the results of Deshawn’s simulation.
T5
T2
H1
T5
H2
T3
H6
H4
H4
T1
H6
H1
T6
T2
T4
T3
H3
H4
H4
T1
Find the probability that a student at Greendale Middle School takes both art and Spanish.
A. Start Fraction 3 over 20 End Fraction
B. two-fifths
C. start fraction 7 over 20 end fraction
D. one-tenth
A. Start Fraction 3 over 20 End Fraction
We can use the results of Deshawn's simulation to create a table of frequencies:
| | Art (T) | Not Art (H) |
|--------|---------|------------|
| Spanish (1, 2, 3, 4) | 3 | 2 |
| Not Spanish (5, 6) | 4 | 11 |
There are 10 students total in the simulation, so the probability that a student takes both art and Spanish is the frequency of students who are in the intersection of the "Art" and "Spanish" categories (i.e. have outcomes T1, T2, T3, or T4), divided by the total number of students:
P(Art and Spanish) = 3/10 = Start Fraction 3 over 20 End Fraction
Therefore, the probability that a student at Greendale Middle School takes both art and Spanish is Start Fraction 3 over 20 End Fraction.
One-half of the students at Greendale Middle School take art. Two-thirds of the students take Spanish. Deshawn uses a coin and a standard number cube as a simulation of the students at Greendale Middle School.
Deshawn lets “T” represent a student taking art and “H” represent a student not taking art. He lets the numbers 1, 2, 3, and 4 on the number cube represent a student taking Spanish, and the numbers 5 and 6 represent a student not taking Spanish. The table shows the results of Deshawn’s simulation.
T5
T2
H1
T5
H2
T3
H6
H4
H4
T1
H6
H1
T6
T2
T4
T3
H3
H4
H4
T1
Find the probability that a student at Greendale Middle School takes neither art nor Spanish.
A. Start Fraction 3 over 20 End Fraction
B. two-fifths
C. start fraction 7 over 20 end fraction
D. one-tenth
B. two-fifths
We can use the results of Deshawn's simulation to create a table of frequencies:
| | Art (T) | Not Art (H) |
|--------|---------|------------|
| Spanish (1, 2, 3, 4) | 3 | 2 |
| Not Spanish (5, 6) | 4 | 11 |
There are 10 students total in the simulation, so the probability that a student takes neither art nor Spanish is the frequency of students in the category "Not Art" and "Not Spanish" (i.e. have outcomes H5 or H6), divided by the total number of students:
P(Not Art and Not Spanish) = (4 + 7)/10 = 11/10
Wait a minute! That can't be right. The probability of any event should be between 0 and 1, inclusive. What went wrong?
It looks like we made a mistake in the calculation. The numerator of our fraction is greater than the denominator, which means our answer is greater than 1. This can't be right! The problem is that we included some outcomes that should not have been counted. Specifically, we included the outcome H4, which corresponds to a student who does not take Spanish but may or may not take art. We need to exclude this outcome from our count.
If we ignore the outcome H4, we get the following table of frequencies:
| | Art (T) | Not Art (H) |
|--------|---------|------------|
| Spanish (1, 2, 3, 4) | 3 | 2 |
| Not Spanish (5, 6) | 4 | 10 |
Now, there are 9 students total in the simulation (excluding H4), so the probability that a student takes neither art nor Spanish is the frequency of students in the category "Not Art" and "Not Spanish" (i.e. have outcomes H5 or H6), divided by the total number of students:
P(Not Art and Not Spanish) = (4 + 10)/9 = 14/9
Once again, we have a problem! The numerator is greater than the denominator, so our answer is greater than 1. It turns out that we still haven't excluded all the unwanted outcomes. Specifically, we need to exclude the outcomes H4 and T4 from our count, because these outcomes correspond to students who take Spanish but not art, and we only want to count students who take neither art nor Spanish.
If we exclude H4 and T4 from our count, we get the following table of frequencies:
| | Art (T) | Not Art (H) |
|--------|---------|------------|
| Spanish (1, 2, 3) | 3 | 1 |
| Not Spanish (5, 6) | 4 | 8 |
Now, there are 8 students total in the simulation (excluding H4 and T4), so the probability that a student takes neither art nor Spanish is the frequency of students in the category "Not Art" and "Not Spanish" (i.e. have outcomes H5 or H6), divided by the total number of students:
P(Not Art and Not Spanish) = 8/8 = 1
Ah, that's better! The probability of a student taking neither art nor Spanish is 1.
Use the following information to solve the problem.
At a summer camp, 20% of the students come from out of state. Megan wants to simulate cabins of 8 campers. She generates random digits from 0 to 9 and lets the digits 0 and 1 represent a camper from out of state. The table below shows her results.
69531217
89542756
89001254
01346895
12468503
20312346
79564328
59868542
75891003
56103249
58630126
57498510
76134860
52974168
03164985
03289642
68533236
94210145
70215789
32605254
What is the experimental probability that in a cabin of 8 campers, at least 2 are from out of state?
A. start fraction 7 over 20 end fraction
B. Start Fraction 13 over 20 End Fraction
C. start fraction 7 over 10 end fraction
D. four-fifths
B. Start Fraction 13 over 20 End Fraction
To find the experimental probability that in a cabin of 8 campers, at least 2 are from out of state, we need to count the number of outcomes in Megan's simulation that have at least 2 of the digits 0 and 1, and divide by the total number of outcomes.
We can do this by counting the number of outcomes that have 0, 1, or 2 of the digits 0 and 1, and subtracting this number from the total number of outcomes (which is 20):
- Number of outcomes with 0, 1, or 2 of the digits 0 and 1: From the table of results, we can see that there are 3 outcomes with 0 of the digits 0 and 1, and 7 outcomes with 1 of the digits 0 and 1. There are no outcomes with 2 or more of the digits 0 and 1. Therefore, the total number of outcomes with 0, 1, or 2 of the digits 0 and 1 is 3 + 7 = 10.
- Total number of outcomes: From the table of results, we can see that there are 20 outcomes in total.
Therefore, the number of outcomes with at least 2 of the digits 0 and 1 is 20 - 10 = 10.
The experimental probability of getting at least 2 campers from out of state in a cabin of 8 campers is therefore 10/20, or Start Fraction 1 over 2 End Fraction.
But wait! We made a mistake. We assumed that each digit represents a camper, but the problem tells us that 20% of the campers are from out of state. This means that only 20% of the digits in Megan's simulation should represent campers from out of state. We need to adjust our calculations accordingly.
There are a total of 16 digits in each outcome, so we expect 0.20 x 16 = 3.2, or about 3, of the digits in each outcome to represent campers from out of state.
Now we can repeat our calculations from before, but counting the number of outcomes with 0, 1, or 2 of the 3 digits that represent campers from out of state:
- Number of outcomes with 0, 1, or 2 of the 3 digits that represent campers from out of state: From the table of results, we can see that there are 3 outcomes with 0 of the 3 digits that represent campers from out of state, and 6 outcomes with
At a summer camp, 20% of the students come from out of state. Megan wants to simulate cabins of 8 campers. She generates random digits from 0 to 9 and lets the digits 0 and 1 represent a camper from out of state. The table below shows her results.
69531217
89542756
89001254
01346895
12468503
20312346
79564328
59868542
75891003
56103249
58630126
57498510
76134860
52974168
03164985
03289642
68533236
94210145
70215789
32605254
What is the experimental probability that in a cabin of 8 campers, none is from out of state?
A. Fraction is 1 over 20
B. one-tenth
C. one-fifth
D. one-fourth