1. You roll a number cube numbered one to six 12 times. P(5) = . What type of probability is illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
2. A jar contains blue, red, yellow, and green marbles. P(blue) = . What type of probability is illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
41 answers
theoretical; the result is based on the number of possible outcomes, as the probability is based on the ratio of blue marbles to the total number of marbles in the jar.
1. You roll a number cube numbered one to six 12 times. P(5) = . What type of probability is
illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes, as the probability is based on the theoretical likelihood of rolling a 5 on a fair 6-sided die. The experiment is being repeated to gather data and confirm the accuracy of the theoretical probability.
2. A jar contains blue, red, yellow, and green marbles. P(blue) = . What type of probability is
illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
illustrated and why?
(1 point)
experimental; the result is based on the number of possible outcomes
experimental; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes
theoretical; the result is found by repeating an experiment
theoretical; the result is based on the number of possible outcomes, as the probability is based on the theoretical likelihood of selecting a blue marble from the jar based on the ratio of blue marbles to the total number of marbles in the jar. No experiment is being repeated to determine the probability in this case, it is calculated based on the theoretical concept of probability.
A number cube is rolled 160 times. The number 2 comes up 39 times. What is the experimental probability of rolling a 2? What is the theoretical probability of rolling a 2?
(1 point)
39/160:1/80
(1 point)
39/160:1/80
The experimental probability of rolling a 2 is 39/160.
The theoretical probability of rolling a 2 on a fair number cube is 1/6, or approximately 0.1667.
The theoretical probability of rolling a 2 on a fair number cube is 1/6, or approximately 0.1667.
A spinner is divided into 11 equal sections numbered from 0 to 10. You spin the spinner once.
What is P(even)?
What is P(even)?
There are 6 even numbers (0, 2, 4, 6, 8, 10) out of a total of 11 possible outcomes on the spinner.
Therefore, the probability of spinning an even number is:
P(even) = number of even outcomes / total number of outcomes
P(even) = 6 / 11
So, the probability of spinning an even number is 6/11.
Therefore, the probability of spinning an even number is:
P(even) = number of even outcomes / total number of outcomes
P(even) = 6 / 11
So, the probability of spinning an even number is 6/11.
A bag contains 7 green marbles and 4 white marbles. You select a marble at random. What are
the odds in favor of picking a green marble?
(1 point)
7:11
7:4
4:7
3:7
the odds in favor of picking a green marble?
(1 point)
7:11
7:4
4:7
3:7
The odds in favor of picking a green marble are:
number of green marbles : number of non-green marbles
= 7 : 4
Therefore, the odds in favor of picking a green marble are 7:4.
number of green marbles : number of non-green marbles
= 7 : 4
Therefore, the odds in favor of picking a green marble are 7:4.
Food Express is running a special promotion in which customers can win a free gallon of milk
with their food purchase if there is a star on their receipt. So far, 147 of the first 156 customers
have not received a star on their receipts. What is the experimental probability of winning a
free gallon of milk?
with their food purchase if there is a star on their receipt. So far, 147 of the first 156 customers
have not received a star on their receipts. What is the experimental probability of winning a
free gallon of milk?
The experimental probability of winning a free gallon of milk can be calculated as:
number of customers who received a star on their receipt / total number of customers
Since the problem does not provide the number of customers who received a star, we will need to calculate that by finding the number of customers who did not receive a star and then subtracting that from the total number of customers:
number of customers who did not receive a star = 147
total number of customers = 156
number of customers who received a star = total number of customers - number of customers who did not receive a star
number of customers who received a star = 156 - 147
number of customers who received a star = 9
Now we can calculate the experimental probability:
number of customers who received a star on their receipt / total number of customers
9 / 156
So, the experimental probability of winning a free gallon of milk is approximately 0.0577 or 5.77%.
number of customers who received a star on their receipt / total number of customers
Since the problem does not provide the number of customers who received a star, we will need to calculate that by finding the number of customers who did not receive a star and then subtracting that from the total number of customers:
number of customers who did not receive a star = 147
total number of customers = 156
number of customers who received a star = total number of customers - number of customers who did not receive a star
number of customers who received a star = 156 - 147
number of customers who received a star = 9
Now we can calculate the experimental probability:
number of customers who received a star on their receipt / total number of customers
9 / 156
So, the experimental probability of winning a free gallon of milk is approximately 0.0577 or 5.77%.
Food Express is running a special promotion in which customers can win a free gallon of milk
with their food purchase if there is a star on their receipt. So far, 147 of the first 156 customers
have not received a star on their receipts. What is the experimental probability of winning a
free gallon of milk?
11/156
49/52
2/39
3/52
with their food purchase if there is a star on their receipt. So far, 147 of the first 156 customers
have not received a star on their receipts. What is the experimental probability of winning a
free gallon of milk?
11/156
49/52
2/39
3/52
The correct answer is:
Experimental probability = number of customers who received a star on their receipt / total number of customers
Since we know that 147 of the first 156 customers have not received a star on their receipts, we can find the number of customers who did receive a star by subtracting:
Total number of customers - Number of customers who did not receive a star = Number of customers who received a star
156 - 147 = 9
So, out of the 156 customers, only 9 received a star on their receipt and are eligible for a free gallon of milk.
Therefore, the experimental probability of winning a free gallon of milk is:
Number of customers who received a star on their receipt / Total number of customers
9 / 156
Which simplifies to approximately:
Experimental probability = 0.0577 or 5.77%
Experimental probability = number of customers who received a star on their receipt / total number of customers
Since we know that 147 of the first 156 customers have not received a star on their receipts, we can find the number of customers who did receive a star by subtracting:
Total number of customers - Number of customers who did not receive a star = Number of customers who received a star
156 - 147 = 9
So, out of the 156 customers, only 9 received a star on their receipt and are eligible for a free gallon of milk.
Therefore, the experimental probability of winning a free gallon of milk is:
Number of customers who received a star on their receipt / Total number of customers
9 / 156
Which simplifies to approximately:
Experimental probability = 0.0577 or 5.77%
A bag contains 7 green marbles, 9 red marbles, 10 orange marbles, 5 brown marbles, and 10
blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
Since the marbles are being replaced after each draw, the outcomes of the two draws are independent.
Therefore, we can find the probability of drawing a red marble on the first draw (P(red)) and the probability of drawing a blue marble on the second draw (P(blue)) and then multiply these probabilities together, using the multiplication rule of probability.
P(red) = number of red marbles / total number of marbles
P(red) = 9 / 41
P(blue) = number of blue marbles / total number of marbles
P(blue) = 10 / 41
P(red, then blue) = P(red) x P(blue)
P(red, then blue) = (9 / 41) x (10 / 41)
P(red, then blue) = 90 / 1681
So, the probability of selecting a red marble on the first draw and a blue marble on the second draw is 90/1681.
Therefore, we can find the probability of drawing a red marble on the first draw (P(red)) and the probability of drawing a blue marble on the second draw (P(blue)) and then multiply these probabilities together, using the multiplication rule of probability.
P(red) = number of red marbles / total number of marbles
P(red) = 9 / 41
P(blue) = number of blue marbles / total number of marbles
P(blue) = 10 / 41
P(red, then blue) = P(red) x P(blue)
P(red, then blue) = (9 / 41) x (10 / 41)
P(red, then blue) = 90 / 1681
So, the probability of selecting a red marble on the first draw and a blue marble on the second draw is 90/1681.
Each of two urns contains green balls and red balls. Urn I contains 10 green balls and 8 red
balls. Urn II contains 3 green balls and 10 red balls. If a ball is drawn from each urn, what is
P(red and red)?
balls. Urn II contains 3 green balls and 10 red balls. If a ball is drawn from each urn, what is
P(red and red)?
We can apply the multiplication rule of probability to find the probability of drawing a red ball from each urn and multiplying the individual probabilities together.
P(red from Urn I) = number of red balls in Urn I / total number of balls in Urn I
P(red from Urn I) = 8 / 18
P(red from Urn I) = 4 / 9
P(red from Urn II) = number of red balls in Urn II / total number of balls in Urn II
P(red from Urn II) = 10 / 13
Since the balls are drawn independently from each urn, we can use the multiplication rule to calculate the probability of drawing a red ball from each urn:
P(red and red) = P(red from Urn I) x P(red from Urn II)
P(red and red) = (4 / 9) x (10 / 13)
P(red and red) = 40 / 117
So, the probability of drawing a red ball from each urn is 40/117.
P(red from Urn I) = number of red balls in Urn I / total number of balls in Urn I
P(red from Urn I) = 8 / 18
P(red from Urn I) = 4 / 9
P(red from Urn II) = number of red balls in Urn II / total number of balls in Urn II
P(red from Urn II) = 10 / 13
Since the balls are drawn independently from each urn, we can use the multiplication rule to calculate the probability of drawing a red ball from each urn:
P(red and red) = P(red from Urn I) x P(red from Urn II)
P(red and red) = (4 / 9) x (10 / 13)
P(red and red) = 40 / 117
So, the probability of drawing a red ball from each urn is 40/117.
If you spin the spinner below twice, what is P(vowel, then K)?
It is not clear what spinner is being referred to, since no spinner is shown. Please provide the necessary information or image to accurately answer the question.
You have six $1 bills, eight $5 bills, two $10 bills, and four $20 bills in your wallet. You select
a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
The probability of selecting a $1 bill, and then a $10 bill without replacement can be found by applying the multiplication rule of probability:
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
To find the probability of selecting a $1 bill on the first draw, we divide the number of $1 bills by the total number of bills:
P($1) = 6 / 20
To find the probability of selecting a $10 bill on the second draw, given that a $1 bill was drawn on the first draw, we divide the number of $10 bills by the remaining bills in the wallet:
P($10 | $1 was already drawn) = 2 / 19
Since one bill was removed from the wallet before the second draw, there are only 19 bills remaining.
Now we can multiply the two probabilities together:
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
P($1, then $10) = (6 / 20) x (2 / 19)
P($1, then $10) = 3 / 190
Therefore, the probability of selecting a $1 bill and then a $10 bill without replacement is 3/190.
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
To find the probability of selecting a $1 bill on the first draw, we divide the number of $1 bills by the total number of bills:
P($1) = 6 / 20
To find the probability of selecting a $10 bill on the second draw, given that a $1 bill was drawn on the first draw, we divide the number of $10 bills by the remaining bills in the wallet:
P($10 | $1 was already drawn) = 2 / 19
Since one bill was removed from the wallet before the second draw, there are only 19 bills remaining.
Now we can multiply the two probabilities together:
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
P($1, then $10) = (6 / 20) x (2 / 19)
P($1, then $10) = 3 / 190
Therefore, the probability of selecting a $1 bill and then a $10 bill without replacement is 3/190.
You have six $1 bills, eight $5 bills, two $10 bills, and four $20 bills in your wallet. You select
a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
77/190
3/100
9/95
2/5
a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
77/190
3/100
9/95
2/5
The correct answer is:
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
where P($1) is the probability of selecting a $1 bill on the first draw and P($10 | $1 was already drawn) is the probability of selecting a $10 bill on the second draw, given that a $1 bill was drawn on the first draw.
To find P($1), we divide the number of $1 bills by the total number of bills:
P($1) = 6 / 20
To find P($10 | $1 was already drawn), we divide the number of $10 bills by the remaining number of bills, after a $1 bill was drawn:
P($10 | $1 was already drawn) = 2 / 19
Since there are only 19 bills left in the wallet after the first draw.
Now, we can calculate the product of these probabilities:
P($1, then $10) = (6 / 20) x (2 / 19)
P($1, then $10) = 6 / 190
P($1, then $10) = 3 / 95
Therefore, the probability of selecting a $1 bill on the first draw and a $10 bill on the second draw without replacement is 3/95.
P($1, then $10) = P($1) x P($10 | $1 was already drawn)
where P($1) is the probability of selecting a $1 bill on the first draw and P($10 | $1 was already drawn) is the probability of selecting a $10 bill on the second draw, given that a $1 bill was drawn on the first draw.
To find P($1), we divide the number of $1 bills by the total number of bills:
P($1) = 6 / 20
To find P($10 | $1 was already drawn), we divide the number of $10 bills by the remaining number of bills, after a $1 bill was drawn:
P($10 | $1 was already drawn) = 2 / 19
Since there are only 19 bills left in the wallet after the first draw.
Now, we can calculate the product of these probabilities:
P($1, then $10) = (6 / 20) x (2 / 19)
P($1, then $10) = 6 / 190
P($1, then $10) = 3 / 95
Therefore, the probability of selecting a $1 bill on the first draw and a $10 bill on the second draw without replacement is 3/95.
A basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5
peaches. Jack picks a fruit at random and does not replace it. Then Bethany picks a fruit at
random. What is the probability that Jack gets a peach and Bethany gets an orange?
peaches. Jack picks a fruit at random and does not replace it. Then Bethany picks a fruit at
random. What is the probability that Jack gets a peach and Bethany gets an orange?
The probability of Jack picking a peach on the first draw is 5/14, since there are 5 peaches and 14 total pieces of fruit in the basket.
Since Jack does not replace the fruit, there are only 13 pieces of fruit left in the basket for Bethany's draw. The probability of Bethany picking an orange from the remaining fruit is 2/13, since there are 2 oranges left.
To find the probability of both events happening together, we multiply their individual probabilities:
P(Jack gets peach, Bethany gets orange) = P(Jack gets peach) × P(Bethany gets orange)
P(Jack gets peach, Bethany gets orange) = (5/14) × (2/13)
P(Jack gets peach, Bethany gets orange) = 10/182
P(Jack gets peach, Bethany gets orange) = 5/91
Therefore, the probability of Jack picking a peach and Bethany picking an orange is 5/91.
Since Jack does not replace the fruit, there are only 13 pieces of fruit left in the basket for Bethany's draw. The probability of Bethany picking an orange from the remaining fruit is 2/13, since there are 2 oranges left.
To find the probability of both events happening together, we multiply their individual probabilities:
P(Jack gets peach, Bethany gets orange) = P(Jack gets peach) × P(Bethany gets orange)
P(Jack gets peach, Bethany gets orange) = (5/14) × (2/13)
P(Jack gets peach, Bethany gets orange) = 10/182
P(Jack gets peach, Bethany gets orange) = 5/91
Therefore, the probability of Jack picking a peach and Bethany picking an orange is 5/91.
The probability of a certain baseball player hitting a foul ball is . How many foul balls
would you expect her to hit after 80 swings?
(1 point)
4
20
40
60
would you expect her to hit after 80 swings?
(1 point)
4
20
40
60
If the probability of the baseball player hitting a foul ball on each swing is 0.35, then the expected number of foul balls she would hit after 80 swings can be found by multiplying the probability by the number of swings:
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.35 × 80
Expected number of foul balls = 28
Therefore, we would expect the player to hit 28 foul balls after 80 swings.
Answer: 28.
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.35 × 80
Expected number of foul balls = 28
Therefore, we would expect the player to hit 28 foul balls after 80 swings.
Answer: 28.
he probability of a certain baseball player hitting a foul ball is . How many foul balls would you expect her to hit after 80 swings?
(1 point)
4
20
40
60
(1 point)
4
20
40
60
The correct answer is:
If the probability of hitting a foul ball on each swing is 0.4 (not given in the options), then the expected number of foul balls the player would hit after 80 swings can be found by multiplying the probability by the number of swings:
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.4 × 80
Expected number of foul balls = 32
Therefore, we would expect the player to hit 32 foul balls after 80 swings.
Answer: 32.
If the probability of hitting a foul ball on each swing is 0.4 (not given in the options), then the expected number of foul balls the player would hit after 80 swings can be found by multiplying the probability by the number of swings:
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.4 × 80
Expected number of foul balls = 32
Therefore, we would expect the player to hit 32 foul balls after 80 swings.
Answer: 32.
The probability of a certain baseball player hitting a foul ball is . How many foul balls would you expect her to hit after 80 swings?
(1 point)
4
20
40
60
(1 point)
4
20
40
60
I apologize for the error in the previous response. The correct answer is:
If the probability of hitting a foul ball on each swing is 0.35, then the expected number of foul balls the player would hit after 80 swings can be found by multiplying the probability by the number of swings:
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.35 × 80
Expected number of foul balls = 28
Therefore, we would expect the player to hit 28 foul balls after 80 swings.
Answer: 28.
If the probability of hitting a foul ball on each swing is 0.35, then the expected number of foul balls the player would hit after 80 swings can be found by multiplying the probability by the number of swings:
Expected number of foul balls = probability of hitting a foul ball × number of swings
Expected number of foul balls = 0.35 × 80
Expected number of foul balls = 28
Therefore, we would expect the player to hit 28 foul balls after 80 swings.
Answer: 28.
A true-false test has 8 questions. What is the probability of guessing the correct answers to all
of the questions?
of the questions?
The probability of guessing the correct answer to any single true-false question is 1/2 or 0.5.
Since guessing the correct answer on one question is independent of guessing the correct answer on any other question, we can use the multiplication rule of probability to find the probability of guessing the correct answer to all of the questions:
Probability of guessing all 8 answers correctly = (probability of guessing one answer correctly) ^8
Probability of guessing all 8 answers correctly = (0.5) ^8
Probability of guessing all 8 answers correctly = 0.00390625 or approximately 0.39%
Therefore, the probability of guessing the correct answers to all 8 questions on a true-false test is approximately 0.39%.
Since guessing the correct answer on one question is independent of guessing the correct answer on any other question, we can use the multiplication rule of probability to find the probability of guessing the correct answer to all of the questions:
Probability of guessing all 8 answers correctly = (probability of guessing one answer correctly) ^8
Probability of guessing all 8 answers correctly = (0.5) ^8
Probability of guessing all 8 answers correctly = 0.00390625 or approximately 0.39%
Therefore, the probability of guessing the correct answers to all 8 questions on a true-false test is approximately 0.39%.
You and 3 friends go to a concert. In how many different ways can you sit in the assigned
seats?
(1 point)
6 ways
12 ways
24 ways
10 way
seats?
(1 point)
6 ways
12 ways
24 ways
10 way
There are 4 people, including yourself, who can potentially sit in 4 seats. The first person has 4 options for where to sit, the second person has 3 options for where to sit, the third person has 2 options for where to sit, and the fourth person has 1 option for where to sit.
Using the multiplication rule of counting, we can find the total number of different ways the 4 people can sit in the assigned seats:
Total number of ways = 4 x 3 x 2 x 1
Total number of ways = 24
Therefore, there are 24 different ways the 4 people can sit in the assigned seats.
Answer: 24 ways.
Using the multiplication rule of counting, we can find the total number of different ways the 4 people can sit in the assigned seats:
Total number of ways = 4 x 3 x 2 x 1
Total number of ways = 24
Therefore, there are 24 different ways the 4 people can sit in the assigned seats.
Answer: 24 ways.
You own 6 pairs of jeans and want to take 2 of them with you on vacation. In how many ways
can you choose 2 pairs of jeans?
can you choose 2 pairs of jeans?
Since order doesn't matter when selecting two pairs of jeans out of six, we can use the combination formula to calculate the number of ways to choose 2 pairs from 6:
The number of combinations of n objects taken r at a time is given by:
nCr = n! / r!(n-r)!
where n is the total number of objects, r is the number of objects to choose.
In this case, we want to choose 2 pairs of jeans out of a total of 6 pairs, so n=6 and r=2.
So, the number of ways to choose 2 pairs of jeans, denoted as 6C2, is:
6C2 = 6! / 2
The number of combinations of n objects taken r at a time is given by:
nCr = n! / r!(n-r)!
where n is the total number of objects, r is the number of objects to choose.
In this case, we want to choose 2 pairs of jeans out of a total of 6 pairs, so n=6 and r=2.
So, the number of ways to choose 2 pairs of jeans, denoted as 6C2, is:
6C2 = 6! / 2
You own 6 pairs of jeans and want to take 2 of them with you on vacation. In how many ways can you choose 2 pairs of jeans?
10
12
15
20
10
12
15
20
You own 6 pairs of jeans and want to take 2 of them with you on vacation. In how many ways can you choose 2 pairs of jeans?
10
12
15
20
10
12
15
20
3 answers