To find the long-run relative frequency of tails, we need to subtract the observed frequency of heads from the total number of tosses and then divide by the total number of tosses.
Total number of tosses: 300
Observed frequency of heads: 180
Number of tails = Total number of tosses - Observed frequency of heads
Number of tails = 300 - 180 = 120
Long-run relative frequency of tails = Number of tails / Total number of tosses
Long-run relative frequency of tails = 120 / 300 = 0.4
Therefore, the long-run relative frequency of tails is 0.4.
Question
A single coin is tossed 300 times. Heads were observed 180 times. What is the long-run relative frequency of tails? Express the answer in decimal form.
11 answers
Probability Unit Test
9 of 159 of 15 Items
Question
An experiment involves picking a card from a selection of cards numbered 5, 10, and 15. In equation form, what is the probability model for this experiment? List the values of x in ascending order.(2 points)
f(x)=
, where x=
,
,
9 of 159 of 15 Items
Question
An experiment involves picking a card from a selection of cards numbered 5, 10, and 15. In equation form, what is the probability model for this experiment? List the values of x in ascending order.(2 points)
f(x)=
, where x=
,
,
The probability model for this experiment can be represented by the following equation:
f(x) = 1/3, where x = 5, 10, 15
This equation states that the probability of picking any of the cards numbered 5, 10, or 15 is equal to 1/3, as there are a total of 3 cards to choose from and each card has an equal chance of being selected.
The values of x in ascending order are:
x = 5, 10, 15
f(x) = 1/3, where x = 5, 10, 15
This equation states that the probability of picking any of the cards numbered 5, 10, or 15 is equal to 1/3, as there are a total of 3 cards to choose from and each card has an equal chance of being selected.
The values of x in ascending order are:
x = 5, 10, 15
If the table below represents a probability model based on observed frequencies, what is the value of x?
Y C A R E S
P(Y) 0.25 0.18 0.30 0.17 x(1 point)
Responses
0.01
0.01
0.20
0.20
0.10
0.10
1
Y C A R E S
P(Y) 0.25 0.18 0.30 0.17 x(1 point)
Responses
0.01
0.01
0.20
0.20
0.10
0.10
1
To find the value of x, we need to sum the probabilities of all outcomes and set it equal to 1 since the sum of all probabilities in a probability model should equal 1:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Now, solve for x:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
0.90 + x = 1
x = 1 - 0.90
x = 0.10
Therefore, the value of x is 0.10.
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Now, solve for x:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
0.90 + x = 1
x = 1 - 0.90
x = 0.10
Therefore, the value of x is 0.10.
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
(1 point)
Responses
50
50
300
300
200
200
100
100
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly selected a ball from both bowls and recorded the results in a table. If there are 500 balls in Bowl B, what is the estimated difference in the expected number of white and red balls in Bowl B?
(1 point)
Responses
50
50
300
300
200
200
100
100
To find the estimated difference in the expected number of white and red balls in Bowl B, we first need to calculate the expected number of white and red balls in Bowl B.
From the table, we see that 60 white balls were selected from Bowl B. If Bowl B contains a total of 500 balls, the estimated expected number of white balls in Bowl B is:
Expected number of white balls in Bowl B = (Number of white balls selected from Bowl B) / (Total number of balls in Bowl B) * 500
Expected number of white balls in Bowl B = (60/100) * 500
Expected number of white balls in Bowl B = 300
Similarly, the estimated expected number of red balls in Bowl B can be calculated as:
Expected number of red balls in Bowl B = (Number of red balls selected from Bowl B) / (Total number of balls in Bowl B) * 500
Expected number of red balls in Bowl B = (40/100) * 500
Expected number of red balls in Bowl B = 200
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is:
300 (expected number of white balls) - 200 (expected number of red balls) = 100
The estimated difference in the expected number of white and red balls in Bowl B is 100.
From the table, we see that 60 white balls were selected from Bowl B. If Bowl B contains a total of 500 balls, the estimated expected number of white balls in Bowl B is:
Expected number of white balls in Bowl B = (Number of white balls selected from Bowl B) / (Total number of balls in Bowl B) * 500
Expected number of white balls in Bowl B = (60/100) * 500
Expected number of white balls in Bowl B = 300
Similarly, the estimated expected number of red balls in Bowl B can be calculated as:
Expected number of red balls in Bowl B = (Number of red balls selected from Bowl B) / (Total number of balls in Bowl B) * 500
Expected number of red balls in Bowl B = (40/100) * 500
Expected number of red balls in Bowl B = 200
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is:
300 (expected number of white balls) - 200 (expected number of red balls) = 100
The estimated difference in the expected number of white and red balls in Bowl B is 100.
Question
What is the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll?(1 point)
Responses
1
1
14
Start Fraction 1 over 4 End Fraction
2
2
12
What is the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll?(1 point)
Responses
1
1
14
Start Fraction 1 over 4 End Fraction
2
2
12
To find the probability of rolling an odd number on the first roll and an even number on the second roll of a six-sided die, we need to consider the total number of outcomes and the number of favorable outcomes.
Odd numbers on a six-sided die are 1, 3, and 5.
Even numbers on a six-sided die are 2, 4, and 6.
The total number of outcomes on each roll is 6, and the number of favorable outcomes for the first roll (odd number) is 3, and for the second roll (even number) is also 3.
The probability of rolling an odd number on the first roll and an even number on the second roll is:
Probability = (Number of favorable outcomes for odd on first roll / Total outcomes) * (Number of favorable outcomes for even on second roll / Total outcomes)
Probability = (3/6) * (3/6)
Probability = (1/2) * (1/2)
Probability = 1/4
Therefore, the probability of rolling an odd number on the first roll and an even number on the second roll is 1/4.
Odd numbers on a six-sided die are 1, 3, and 5.
Even numbers on a six-sided die are 2, 4, and 6.
The total number of outcomes on each roll is 6, and the number of favorable outcomes for the first roll (odd number) is 3, and for the second roll (even number) is also 3.
The probability of rolling an odd number on the first roll and an even number on the second roll is:
Probability = (Number of favorable outcomes for odd on first roll / Total outcomes) * (Number of favorable outcomes for even on second roll / Total outcomes)
Probability = (3/6) * (3/6)
Probability = (1/2) * (1/2)
Probability = 1/4
Therefore, the probability of rolling an odd number on the first roll and an even number on the second roll is 1/4.
Probability Unit Test
13 of 1513 of 15 Items
Question
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.(1 point)
13 of 1513 of 15 Items
Question
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.(1 point)
To find the probability of selecting two carnations from the big flower vase without replacement, we need to calculate the total number of ways to choose 2 flowers out of all the flowers in the vase, as well as the number of ways to choose 2 carnations.
Total number of flowers in the vase = 5 begonias + 6 carnations + 7 tulips = 18 flowers
To find the number of ways to choose 2 carnations out of 6 carnations, we use the combination formula:
Number of ways to choose 2 carnations = C(6, 2) = 6! / (2! * (6-2)!) = 15 ways
Similarly, the total number of ways to choose 2 flowers out of 18 flowers is:
Total number of ways to choose 2 flowers = C(18, 2) = 18! / (2! * (18-2)!) = 153 ways
The probability of selecting two carnations is:
Probability = Number of ways to choose 2 carnations / Total number of ways to choose 2 flowers
Probability = 15 / 153
Probability = 5 / 51
Therefore, the probability of selecting two carnations from the vase is 5/51.
Total number of flowers in the vase = 5 begonias + 6 carnations + 7 tulips = 18 flowers
To find the number of ways to choose 2 carnations out of 6 carnations, we use the combination formula:
Number of ways to choose 2 carnations = C(6, 2) = 6! / (2! * (6-2)!) = 15 ways
Similarly, the total number of ways to choose 2 flowers out of 18 flowers is:
Total number of ways to choose 2 flowers = C(18, 2) = 18! / (2! * (18-2)!) = 153 ways
The probability of selecting two carnations is:
Probability = Number of ways to choose 2 carnations / Total number of ways to choose 2 flowers
Probability = 15 / 153
Probability = 5 / 51
Therefore, the probability of selecting two carnations from the vase is 5/51.