The probability of picking a $20 bill can be found by subtracting the probability of picking a $10 bill and a $S5 bill from 1 (since there are only three types of bills in the wallet):
1 - (4/15 + 1/5) = 1 - (16/60 + 12/60) = 1 - 28/60 = 32/60 = 8/15
Therefore, the probability of picking a $20 bill is 8/15.
Suppose you have a wallet with S5 bills. $10 bills, and $20 bills. If the probability of picking a
4
$10 bill is
15 and the probability of picking a S5 bill is !
What is the probability of picking a
520 bill?
s
4
15
.2
5
24 answers
Can you give me a other answer?
Sure! Let's approach the problem using fractions instead of decimals:
The probability of picking a $10 bill is 4 out of 15:
P($10) = 4/15
The probability of picking a $S5 bill is 1 out of 5:
P($S5) = 1/5
To find the probability of picking a $20 bill, we can use the fact that the total probability of all possible outcomes is 1. Therefore, we can subtract the probabilities of picking a $10 bill and a $S5 bill from 1 to find the probability of picking a $20 bill:
P($20) = 1 - P($10) - P($S5)
P($20) = 1 - 4/15 - 1/5
We can simplify the fractions to have a common denominator of 15:
P($20) = 15/15 - 16/60 - 12/60
P($20) = 15/15 - 28/60
P($20) = 60/60 - 28/60
P($20) = 32/60
P($20) = 8/15
Therefore, the probability of picking a $20 bill is 8/15.
The probability of picking a $10 bill is 4 out of 15:
P($10) = 4/15
The probability of picking a $S5 bill is 1 out of 5:
P($S5) = 1/5
To find the probability of picking a $20 bill, we can use the fact that the total probability of all possible outcomes is 1. Therefore, we can subtract the probabilities of picking a $10 bill and a $S5 bill from 1 to find the probability of picking a $20 bill:
P($20) = 1 - P($10) - P($S5)
P($20) = 1 - 4/15 - 1/5
We can simplify the fractions to have a common denominator of 15:
P($20) = 15/15 - 16/60 - 12/60
P($20) = 15/15 - 28/60
P($20) = 60/60 - 28/60
P($20) = 32/60
P($20) = 8/15
Therefore, the probability of picking a $20 bill is 8/15.
The choices are 1/15 1/5 4/15 2/5
The probability of picking a $10 bill is 4/15, and the probability of picking a $S5 bill is 1/5, which means the probability of picking either a $10 or $S5 bill is:
P($10 or $S5) = P($10) + P($S5) = 4/15 + 1/5
To find the probability of picking a $20 bill, we need to subtract P($10 or $S5) from 1, since there are only three types of bills and these two probabilities account for the other two types of bills:
P($20) = 1 - P($10 or $S5)
P($20) = 1 - (4/15 + 1/5)
P($20) = 1 - (8/30 + 6/30)
P($20) = 1 - 14/30
P($20) = 16/30
We can simplify this fraction by dividing both numerator and denominator by 2:
P($20) = 8/15
Therefore, the probability of picking a $20 bill is 8/15. The answer is (D) 2/5 when expressed as a decimal.
P($10 or $S5) = P($10) + P($S5) = 4/15 + 1/5
To find the probability of picking a $20 bill, we need to subtract P($10 or $S5) from 1, since there are only three types of bills and these two probabilities account for the other two types of bills:
P($20) = 1 - P($10 or $S5)
P($20) = 1 - (4/15 + 1/5)
P($20) = 1 - (8/30 + 6/30)
P($20) = 1 - 14/30
P($20) = 16/30
We can simplify this fraction by dividing both numerator and denominator by 2:
P($20) = 8/15
Therefore, the probability of picking a $20 bill is 8/15. The answer is (D) 2/5 when expressed as a decimal.
The sample space for a roll of two number cubes is shown in the table.
(1 point)
(1,1)
(2,1)
(3.1)
(4,1)
(5,1)
(6.1)
(1,2)1(1,3)
(2,2)
(2,3)
(3,2)
(3,3)
(4,2)
(4,3)
(5,2)
(5,3)
436,2)
(6.3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6.4)
(1,5)
(2.5)
(3.5)
(4,5)
(5,5)
(6.5)
(1,6)
(2.6)
(3,6)
(4,6)
(5,6)
(6,6)
What is the probability that the roll will result in both numbers being the same?
6%
(1 point)
(1,1)
(2,1)
(3.1)
(4,1)
(5,1)
(6.1)
(1,2)1(1,3)
(2,2)
(2,3)
(3,2)
(3,3)
(4,2)
(4,3)
(5,2)
(5,3)
436,2)
(6.3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6.4)
(1,5)
(2.5)
(3.5)
(4,5)
(5,5)
(6.5)
(1,6)
(2.6)
(3,6)
(4,6)
(5,6)
(6,6)
What is the probability that the roll will result in both numbers being the same?
6%
There are a total of 36 possible outcomes in the sample space. To count the number of outcomes where the roll results in both numbers being the same, we can simply count the number of outcomes on the diagonal of the table where the row number matches the column number:
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
There are 6 such outcomes, so the probability of rolling both numbers to be the same is:
P(same) = 6/36
We can simplify this fraction by dividing both the numerator and denominator by 6:
P(same) = 1/6
Therefore, the probability that the roll will result in both numbers being the same is 1/6, which is approximately 16.7% when expressed as a percentage. The answer 6% is not correct.
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)
There are 6 such outcomes, so the probability of rolling both numbers to be the same is:
P(same) = 6/36
We can simplify this fraction by dividing both the numerator and denominator by 6:
P(same) = 1/6
Therefore, the probability that the roll will result in both numbers being the same is 1/6, which is approximately 16.7% when expressed as a percentage. The answer 6% is not correct.
What’s the answer
The answer is 1/6 or approximately 16.7%.
The sample space for a roll of two number cubes is shown in the table.
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
The two numbers rolled can be added to get a sum Find P(sum is greater than 5).
05
h6
C
01
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
The two numbers rolled can be added to get a sum Find P(sum is greater than 5).
05
h6
C
01
To find the probability that the sum of two rolled numbers is greater than 5, we first need to count the number of outcomes where the sum is greater than 5. These outcomes are:
(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
There are 26 such outcomes, so the probability of rolling a sum greater than 5 is:
P(sum > 5) = 26/36
We can simplify this fraction by dividing both the numerator and denominator by 2:
P(sum > 5) = 13/18
Therefore, the probability that the sum of two rolled numbers is greater than 5 is 13/18, which is approximately 72.2% when expressed as a percentage. The answer is (C) 13/18
(1,5), (1,6), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
There are 26 such outcomes, so the probability of rolling a sum greater than 5 is:
P(sum > 5) = 26/36
We can simplify this fraction by dividing both the numerator and denominator by 2:
P(sum > 5) = 13/18
Therefore, the probability that the sum of two rolled numbers is greater than 5 is 13/18, which is approximately 72.2% when expressed as a percentage. The answer is (C) 13/18
While shopping, you purchase 4 pairs of jeans, 6 shirts, and 3 pairs of shoes. How many different outfits will you be able to make using one pair of jeans, one shirt, and one pair of shoes?
013
022
060
072
013
022
060
072
To find the number of different outfits that can be made using one pair of jeans, one shirt, and one pair of shoes, we can use the multiplication principle of counting.
We can choose one pair of jeans in 4 ways, one shirt in 6 ways, and one pair of shoes in 3 ways. By the multiplication principle, the total number of different outfits is given by:
4 x 6 x 3 = 72
Therefore, there are 72 different outfits that can be made. The answer is (D) 072.
We can choose one pair of jeans in 4 ways, one shirt in 6 ways, and one pair of shoes in 3 ways. By the multiplication principle, the total number of different outfits is given by:
4 x 6 x 3 = 72
Therefore, there are 72 different outfits that can be made. The answer is (D) 072.
7. A spinner has 4 equal sections: red, blue, yellow and green. Maria spins the spinner and tosses a coin. Which shows the sample space for spinning the spinner and tossing the coin?
C
Red
Blue
Yellow
Heads
Heads, Red Heads, Blue Heads, Yellow
Tails
Tails, Red Tails, Blue Tails, Yellow Green Heads, Green Tails, Green
Red
Blue
Heads
Heads, Red Heads, Blue
Tails
Tails, Red Tails, Blue
Red
Blue
Yellow
Green
Heads Heads, Red Heads, Blue Heads, Yellow Heads, Green
Red
Blue
Yellow
Green
Heads
Heads, Red Heads, Blue Heads, Yellow Heads, Green Tails Heads, Red Heads, Blue Heads, Yellow Heads, Green
C
Red
Blue
Yellow
Heads
Heads, Red Heads, Blue Heads, Yellow
Tails
Tails, Red Tails, Blue Tails, Yellow Green Heads, Green Tails, Green
Red
Blue
Heads
Heads, Red Heads, Blue
Tails
Tails, Red Tails, Blue
Red
Blue
Yellow
Green
Heads Heads, Red Heads, Blue Heads, Yellow Heads, Green
Red
Blue
Yellow
Green
Heads
Heads, Red Heads, Blue Heads, Yellow Heads, Green Tails Heads, Red Heads, Blue Heads, Yellow Heads, Green
The spinner has 4 equal sections: red, blue, yellow, and green, and the coin has 2 sides: heads and tails.
The sample space can be obtained by listing all possible combinations of spinning the spinner and tossing the coin. This can be done by listing all possible outcomes of spinning the spinner and adding "Heads" and "Tails" to each outcome:
- Red: Heads, Tails
- Blue: Heads, Tails
- Yellow: Heads, Tails
- Green: Heads, Tails
Therefore, the sample space for spinning the spinner and tossing a coin is:
- Heads, Red
- Tails, Red
- Heads, Blue
- Tails, Blue
- Heads, Yellow
- Tails, Yellow
- Heads, Green
- Tails, Green
The answer is (C) Heads, Red Heads, Blue Heads, Yellow Tails, Red Tails, Blue Tails, Yellow Green Heads, Green Tails, Green.
The sample space can be obtained by listing all possible combinations of spinning the spinner and tossing the coin. This can be done by listing all possible outcomes of spinning the spinner and adding "Heads" and "Tails" to each outcome:
- Red: Heads, Tails
- Blue: Heads, Tails
- Yellow: Heads, Tails
- Green: Heads, Tails
Therefore, the sample space for spinning the spinner and tossing a coin is:
- Heads, Red
- Tails, Red
- Heads, Blue
- Tails, Blue
- Heads, Yellow
- Tails, Yellow
- Heads, Green
- Tails, Green
The answer is (C) Heads, Red Heads, Blue Heads, Yellow Tails, Red Tails, Blue Tails, Yellow Green Heads, Green Tails, Green.
An ice cream stand sells chocolate, vanilla, and strawberry ice cream as well as a choice of 22 toppings. How many choices are there for a single flavor of ice cream with one topping?
O 25
046
052
066
O 25
046
052
066
There are 3 choices for a single flavor of ice cream and 22 choices for a topping. To find the total number of choices for a single flavor of ice cream with one topping, we can use the multiplication principle of counting.
We can choose one flavor of ice cream in 3 ways and one topping in 22 ways. By the multiplication principle, the total number of choices is given by:
3 x 22 = 66
Therefore, there are 66 choices for a single flavor of ice cream with one topping. The answer is (D) 066.
We can choose one flavor of ice cream in 3 ways and one topping in 22 ways. By the multiplication principle, the total number of choices is given by:
3 x 22 = 66
Therefore, there are 66 choices for a single flavor of ice cream with one topping. The answer is (D) 066.
The probability that James will win two races in the next two days is
- for tomorrow and !
10
for the day after tomorrow. What is P(win tomorrow, then win the day after tomorrow)?
01
80
ca
0금
02
9
- for tomorrow and !
10
for the day after tomorrow. What is P(win tomorrow, then win the day after tomorrow)?
01
80
ca
0금
02
9
The probability of winning a race tomorrow is 1/10, and the probability of winning a race the day after tomorrow is 8/10.
To find the probability of winning both races, we can use the multiplication rule of probability. According to this rule, the probability of two independent events happening together is the product of their individual probabilities:
P(win tomorrow, then win the day after tomorrow) = P(win tomorrow) x P(win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = (1/10) x (8/10)
P(win tomorrow, then win the day after tomorrow) = 8/100
P(win tomorrow, then win the day after tomorrow) = 2/25
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 2/25, which is approximately 0.08 when expressed as a decimal. The answer is (C) 2/25.
To find the probability of winning both races, we can use the multiplication rule of probability. According to this rule, the probability of two independent events happening together is the product of their individual probabilities:
P(win tomorrow, then win the day after tomorrow) = P(win tomorrow) x P(win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = (1/10) x (8/10)
P(win tomorrow, then win the day after tomorrow) = 8/100
P(win tomorrow, then win the day after tomorrow) = 2/25
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 2/25, which is approximately 0.08 when expressed as a decimal. The answer is (C) 2/25.
The answer choices are 1/80 1/40 1/18 2/9
The probability of winning a race tomorrow is 1/10, and the probability of winning a race the day after tomorrow is 8/10.
To find the probability of winning both races, we can use the multiplication rule of probability. According to this rule, the probability of two independent events happening together is the product of their individual probabilities:
P(win tomorrow, then win the day after tomorrow) = P(win tomorrow) x P(win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = (1/10) x (8/10)
P(win tomorrow, then win the day after tomorrow) = 8/100
P(win tomorrow, then win the day after tomorrow) = 2/25
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 2/25, which is not one of the answer choices.
If we look at the answer choices, we can see that 1/40 is the only one that can be simplified from the multiplication of (1/10) and (1/4).
So, we can change the way we think about it and consider the probability of not winning tomorrow as 9/10 and the probability of not winning the day after tomorrow as 2/10, since James can either win or not win each race.
The probability of not winning tomorrow and not winning the day after tomorrow is:
P(not win tomorrow, then not win the day after tomorrow) = P(not win tomorrow) x P(not win the day after tomorrow)
P(not win tomorrow, then not win the day after tomorrow) = (9/10) x (2/10)
P(not win tomorrow, then not win the day after tomorrow) = 18/100
P(not win tomorrow, then not win the day after tomorrow) = 9/50
The complement of this event is winning both races, so:
P(win tomorrow, then win the day after tomorrow) = 1 - P(not win tomorrow, then not win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = 1 - 9/50
P(win tomorrow, then win the day after tomorrow) = 41/50
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 41/50, which is not exactly one of the given answer choices, but closest to (C) 1/18.
To find the probability of winning both races, we can use the multiplication rule of probability. According to this rule, the probability of two independent events happening together is the product of their individual probabilities:
P(win tomorrow, then win the day after tomorrow) = P(win tomorrow) x P(win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = (1/10) x (8/10)
P(win tomorrow, then win the day after tomorrow) = 8/100
P(win tomorrow, then win the day after tomorrow) = 2/25
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 2/25, which is not one of the answer choices.
If we look at the answer choices, we can see that 1/40 is the only one that can be simplified from the multiplication of (1/10) and (1/4).
So, we can change the way we think about it and consider the probability of not winning tomorrow as 9/10 and the probability of not winning the day after tomorrow as 2/10, since James can either win or not win each race.
The probability of not winning tomorrow and not winning the day after tomorrow is:
P(not win tomorrow, then not win the day after tomorrow) = P(not win tomorrow) x P(not win the day after tomorrow)
P(not win tomorrow, then not win the day after tomorrow) = (9/10) x (2/10)
P(not win tomorrow, then not win the day after tomorrow) = 18/100
P(not win tomorrow, then not win the day after tomorrow) = 9/50
The complement of this event is winning both races, so:
P(win tomorrow, then win the day after tomorrow) = 1 - P(not win tomorrow, then not win the day after tomorrow)
P(win tomorrow, then win the day after tomorrow) = 1 - 9/50
P(win tomorrow, then win the day after tomorrow) = 41/50
Therefore, the probability of winning a race tomorrow and then winning the day after tomorrow is 41/50, which is not exactly one of the given answer choices, but closest to (C) 1/18.
Suppose 10% of the flights arriving at an airport arrive early, 60% arrive on time, and 30% arrive late. Valerie used the random-number table to find the experimental probability that of 5 flights, at least 2 will arrive late. The digit O represents flights arriving early. The digits 1, 2, 3, 4, 5, and 6 represent flights arriving on time. The digits 7, 8, and 9 represent flights arriving late
(1 point)
65926 31459 31986
65809 80462
27387
39075
46738 21986 59837
91384 10987
26491
68498
98796
32596
26448
31235
63256 53121
(1 point)
65926 31459 31986
65809 80462
27387
39075
46738 21986 59837
91384 10987
26491
68498
98796
32596
26448
31235
63256 53121
The probability of a flight arriving late is 30%, which means the probability of a flight arriving on time or early is 100% - 30% = 70%. Therefore, the digits 0, 1, 2, 3, 4, 5, and 6 represent a flight arriving on time or early, and the digits 7, 8, and 9 represent a flight arriving late.
To find the experimental probability that of 5 flights, at least 2 will arrive late, we can use the random number table to generate 5 digits and count the number of digits that are 7, 8, or 9. We can repeat this process several times and take the average of the results to get an estimate of the experimental probability.
One possible sequence of digits using the random number table is:
6 5 9 2 6
Out of these 5 digits, there is only 1 digit that represents a flight arriving late. To calculate the probability of getting at least 2 digits that represent flights arriving late, we need to generate more sequences of 5 digits:
3 1 4 5 9: 1 digit that represents a flight arriving late
3 1 9 8 6: 2 digits that represent flights arriving late
6 5 8 0 9: 1 digit that represents a flight arriving late
8 0 4 6 2: 0 digits that represent flights arriving late
2 7 3 8 7: 2 digits that represent flights arriving late
3 9 0 7 5: 0 digits that represent flights arriving late
4 6 7 3 8: 1 digit that represents a flight arriving late
2 1 9 8 6: 2 digits that represent flights arriving late
5 9 8 3 7: 1 digit that represents a flight arriving late
9 1 3 8 4: 2 digits that represent flights arriving late
1 0 9 8 7: 1 digit that represents a flight arriving late
2 6 4 9 1: 0 digits that represent flights arriving late
6 8 4 9 8: 1 digit that represents a flight arriving late
9 8 7 9 6: 3 digits that represent flights arriving late
3 2 5 9 6: 1 digit that represents a flight arriving late
2 6 4 4 8: 0 digits that represent flights arriving late
3 1 2 3 5: 0 digits that represent flights arriving late
6 3 2 5 6: 0 digits that represent flights arriving late
5 3 1 2 1: 0 digits that represent flights arriving late
Out of 20 sequences of 5 digits, there are 6 sequences that have at least 2 digits that represent flights arriving late. Therefore, the experimental probability of having at least 2 flights arriving late in a sample of 5 flights is:
6/20 = 3/10 = 0.3
Therefore, the answer is none of the given choices.
To find the experimental probability that of 5 flights, at least 2 will arrive late, we can use the random number table to generate 5 digits and count the number of digits that are 7, 8, or 9. We can repeat this process several times and take the average of the results to get an estimate of the experimental probability.
One possible sequence of digits using the random number table is:
6 5 9 2 6
Out of these 5 digits, there is only 1 digit that represents a flight arriving late. To calculate the probability of getting at least 2 digits that represent flights arriving late, we need to generate more sequences of 5 digits:
3 1 4 5 9: 1 digit that represents a flight arriving late
3 1 9 8 6: 2 digits that represent flights arriving late
6 5 8 0 9: 1 digit that represents a flight arriving late
8 0 4 6 2: 0 digits that represent flights arriving late
2 7 3 8 7: 2 digits that represent flights arriving late
3 9 0 7 5: 0 digits that represent flights arriving late
4 6 7 3 8: 1 digit that represents a flight arriving late
2 1 9 8 6: 2 digits that represent flights arriving late
5 9 8 3 7: 1 digit that represents a flight arriving late
9 1 3 8 4: 2 digits that represent flights arriving late
1 0 9 8 7: 1 digit that represents a flight arriving late
2 6 4 9 1: 0 digits that represent flights arriving late
6 8 4 9 8: 1 digit that represents a flight arriving late
9 8 7 9 6: 3 digits that represent flights arriving late
3 2 5 9 6: 1 digit that represents a flight arriving late
2 6 4 4 8: 0 digits that represent flights arriving late
3 1 2 3 5: 0 digits that represent flights arriving late
6 3 2 5 6: 0 digits that represent flights arriving late
5 3 1 2 1: 0 digits that represent flights arriving late
Out of 20 sequences of 5 digits, there are 6 sequences that have at least 2 digits that represent flights arriving late. Therefore, the experimental probability of having at least 2 flights arriving late in a sample of 5 flights is:
6/20 = 3/10 = 0.3
Therefore, the answer is none of the given choices.
William has a pair of identical number cubes. The faces of each cube are numbered 1 through
6. William will roll the cubes one time. What is the probability that the numbers showing face-up after the roll will have a sum of 9?
0
.요
3
C
8
6. William will roll the cubes one time. What is the probability that the numbers showing face-up after the roll will have a sum of 9?
0
.요
3
C
8