What are the similarities between organized counting and permutation? Explain using examples.
I can only think of one similarity: they both use probability problems using counting P(event)= Number of favourable outcomes/Total number of possible outcomes.
Thanks for your help.
1. This tree diagram shows the number of theaters and the number of different movies being shown at theater. According to the diagram, how many possible movie choices are there?
A. 2
B.3
C.6
D.8
C.6
Which scenario is represented by this tree diagram?
A. 3 kinds of soup in 4 different sizes
B. 2 pairs of shoes and 5 colors of socks
C. 2 different CDs with 6 songs on each
D. 2 types of cake, in 2 different shapes, with 2 different types of frosting
D. 2 types of cake, in 2 different shapes, with 2 different types of frosting
. Find the total number of possible outcomes for 4 different cell phone models, each of which has 3
A. 7
B. 12
C. 16
D. 24
D. 24
There are 4 different cell phone models and each model has 3 different colors. So, the total number of possible outcomes is:
4 × 3 = 12
And since each phone model can have 2 different memory options:
12 × 2 = 24.
. Find the total number of possible outcomes for 10 styles of running shoes, each in a men's and women's version
A. 8
B. 12
C. 20
D. 100
C. 20
There are 10 styles of running shoes and each style has a men's and women's version. So, the total number of possible outcomes is:
10 × 2 = 20.
A pizza shop offers 12 different pizza toppings. If the total number of possible outcomes is 36, how many different sizes of single topping pizzas do they offer?
A. 2
B. 3
C. 6
D. 24
B. 3
The total number of possible outcomes is given as 36, which means there are 36 different ways to combine toppings on a pizza. We know that there are 12 different toppings, so we can assume that each pizza can have up to three toppings (because 12^3 = 1,728, which is more than 36).
So, the pizza shop offers three different sizes of single topping pizzas.
Which of the following sample spaces correctly lists all the possible outcomes for flipping a coin and rolling a number cube? Let head be H and tail be T
A. н-1, Н-2, H-3, H-4, H-5, H-6
B. H-1, T-1, H-2, I-2, H-3, I-3, H-4, I-4, H-5, I-5, H-6, I-6
C. H-1, H-2, H-3, T-4, I-5, I-6
B. H-1, T-1, H-2, I-2, H-3, I-3, H-4, I-4, H-5, I-5, H-6, I-6
This sample space correctly lists all the possible outcomes for flipping a coin (H for heads, T for tails) and rolling a number cube (with numbers 1-6 represented by 1, 2, 3, 4, 5, and 6). Each outcome consists of a combination of a coin flip and a number rolled on the cube.
. How many different lunch combinations can be made from three sandwich choices, two side item choices, and four beverage choices if you choose one sandwich, one side, and one beverage?
A. 9
B. 20
C. 24
C. 24
To find the total number of lunch combinations, we multiply the number of choices for each category:
3 sandwich choices × 2 side item choices × 4 beverage choices = 24
Therefore, there are 24 different lunch combinations that can be made.
A bicycle manufacturer offers two styles, three sizes, and five different colors. How many different bicycles are offered?
A. 30
B. 25
C. 10
B. 25
To find the total number of different bicycles offered, we multiply the number of options for each category:
2 styles × 3 sizes × 5 colors = 30
However, we are told that there are only two styles, so we need to remove the extra 5 bicycles that are not offered in one of the styles:
30 - 5 = 25
Therefore, there are 25 different bicycles offered.
A restaurant offers four different appetizers, nine different entrees, and three different desserts.
How many different meals can you order of one appetizer, one entrée, and one dessert?
A. 98
B. 108
C. 120
B. 108
To find the total number of different meals, we multiply the number of options for each category:
4 appetizers × 9 entrees × 3 desserts = 108
Therefore, there are 108 different meals that can be ordered.
1. What is the experimental probability of the spinner landing on red?
A. 1/2
B. 1/3
C. 1/4
B. 1/3
We don't have a specific spinner given here, but assuming a standard spinner with three equal sections for red, blue, and green, each with the same chance of being landed on, the experimental probability of the spinner landing on red would be:
Number of times the spinner lands on red ÷ Total number of spins
If we spun the spinner 3 times and landed on red once, the experimental probability would be:
1 ÷ 3 = 1/3
Therefore, the experimental probability of the spinner landing on red is 1/3.
What is the theoretical probability of the spinner landing on red?
A. 1/2
B. 1/3
C. 1/4
B. 1/3
If we assume a standard spinner with three equal sections for red, blue, and green, each with the same chance of being landed on, the theoretical probability of the spinner landing on red would be:
Number of ways the spinner can land on red ÷ Total number of possible outcomes
The spinner can land on red in 1 way, out of a total of 3 possible outcomes (red, blue, or green), so:
1 ÷ 3 = 1/3
Therefore, the theoretical probability of the spinner landing on red is 1/3.
3. What is the experimental probability that at least one of the coins landed on heads? Express the probability as a decimal.
A. 0.8
B. 0.2
C. 0.6
To answer this question, we need to know how many times we flipped the coins and how many times at least one coin landed on heads. Without that information, we cannot give a specific answer.
However, in general, the experimental probability of at least one coin landing on heads when flipping two coins is the same as the probability of getting at least one "success" in two independent trials of a "success-failure" experiment, where success is getting heads on a coin flip and failure is getting tails.
The probability of getting at least one success in two independent trials can be found using the complement rule:
P(at least one success) = 1 - P(no successes)
The probability of getting no successes (both coins landing on tails) is:
P(no successes) = (1/2) x (1/2) = 1/4
So the probability of getting at least one success (at least one coin landing on heads) is:
P(at least one success) = 1 - P(no successes) = 1 - 1/4 = 3/4
So the experimental probability of at least one coin landing on heads when flipping two coins is 3/4 or 0.75 as a decimal.
4. Make a tree diagram for two coins that are tossed. Find the theoretical probability that at least one coin is (1 point heads. Express the probability as a decimal.
A. 0.5
B. 0.75
C. 0.8
Here is the tree diagram for two coins being tossed:
```
H T
/ \ / \
H T H T
```
The theoretical probability of at least one coin landing on heads can be found by adding up the probabilities of the possible outcomes in which at least one coin lands on heads:
- HH: both coins land on heads (probability = 1/4)
- HT: one coin lands on heads and one lands on tails (probability = 1/4)
- TH: one coin lands on tails and one lands on heads (probability = 1/4)
The probability of getting at least one head is the sum of these probabilities:
P(at least one head) = P(HH) + P(HT) + P(TH) = 1/4 + 1/4 + 1/4 = 3/4
So the theoretical probability of getting at least one head when tossing two coins is 3/4 or 0.75 as a decimal.
Therefore, the answer is B. 0.75.
Find the experimental probability, and express it as a percent.
A. 50%
B. 40%
C. 29%
I'm sorry, but the experimental probability was not given in the question. It's impossible to determine the answer without that information.
5. Find the experimental probability, and express it as a percent.
P(even) =?
A. 50%
B. 40%
C. 29%
Without additional information or context, it's impossible to determine the experimental probability of getting an even number. However, assuming we are talking about rolling a standard six-sided die, we can determine the probability of rolling an even number as follows:
There are three even numbers on a standard die: 2, 4, and 6. There are a total of six possible outcomes when rolling a die: 1, 2, 3, 4, 5, or 6. Therefore, the probability of rolling an even number is:
P(even) = Number of even outcomes ÷ Total number of possible outcomes = 3 ÷ 6 = 1/2
So the experimental probability of rolling an even number would be the number of times an even number was rolled divided by the total number of rolls. Without that information, we cannot determine the answer.
But the theoretical probability of getting an even number is 1/2 or 50%, which is the answer A. 50%.
6. How does the experimental probability you found in question 5 compare to the theoretical probability of
rolling an even number?
A. The theoretical probability is 10% greater than the experimental probability.
B. The experimental probability is 10% greater than the theoretical probability.
C. The theoretical probability is equal to the experimental probability.
C. The theoretical probability is equal to the experimental probability.
The theoretical probability of rolling an even number on a standard six-sided die is 1/2 or 50%. This is the probability based on the assumption that each outcome (rolling 1, 2, 3, 4, 5, or 6) is equally likely.
The experimental probability of rolling an even number depends on the number of rolls and the actual outcomes. Without that information, we cannot determine the experimental probability. However, if we assume a large enough number of rolls, the experimental probability should approach the theoretical probability of 50%.
Since the theoretical probability of rolling an even number is 50% and the experimental probability should approach this value as the number of trials increases, it follows that the theoretical and experimental probabilities are equal. Therefore, the answer is C. The theoretical probability is equal to the experimental probability.
Which of the following sample spaces correctly lists all the possible outcomes for flipping a coin and spinning a four-colored spinner (red, blue, green, and yellow)?
A.H-red, H-blue, H-green, H-yellow, T-yellow, T-green, T-blue, T-red
B.H-red ,T-red, H-blue, T-blue, H-green, T-yellow
C.T-yellow, H-red, T-blue, H-green, T-green, H-yellow, T-red
D.H-red, T-blue, H-green, T-yellow
A.H-red, H-blue, H-green, H-yellow, T-yellow, T-green, T-blue, T-red
This sample space correctly lists all the possible outcomes for flipping a coin (H for heads, T for tails) and spinning a four-colored spinner (red, blue, green, and yellow). Each outcome consists of a combination of a coin flip and one of the four spinner colors.
How many different lunch combinations can be made from two sandwich choices, three side item choices, and three beverage choices if you choose one sandwich, one side, and one beverage?
A.8
B.9
C.18
D.11
The number of lunch combinations can be found by multiplying the number of sandwich choices, side item choices, and beverage choices together:
2 sandwich choices x 3 side item choices x 3 beverage choices = 18
Therefore, there are 18 different lunch combinations that can be made.
The answer is C. 18.
A number cube is rolled. What is P(3 or an even number)?
A. 1/2
B. 2/3
C. 4/2
D. 1/6
P(3 or an even number) means the probability of rolling a 3 or an even number.
There are three even numbers on a standard six-sided die: 2, 4, and 6. And there is only one 3 on the die. Therefore, the total number of possible outcomes is 6, and the number of outcomes that satisfy the condition of rolling a 3 or an even number is 3 (2, 4, and 6).
So the probability of rolling a 3 or an even number is:
P(3 or an even number) = Number of outcomes that satisfy the condition ÷ Total number of possible outcomes = 3 ÷ 6 = 1/2
Therefore, the answer is A. 1/2.
Which of the following is a certain event?
A. rolling a composite number on a number cube
B. rolling a number greater than 1 on a number cube
C. rolling an odd or even number on a number cube
D. rolling an odd or even number greater than 7
B. rolling a number greater than 1 on a number cube
A certain event is one that is guaranteed to happen, or has a probability of 1. Rolling a number greater than 1 on a number cube is a certain event because every number on a standard cube is greater than 1, so this event is guaranteed to occur.
The other events listed are not certain events because they have some chance of not occurring. For example, rolling a composite number on a number cube (A) has a probability of 4/6 or 2/3, and rolling an odd or even number greater than 7 (D) has a probability of 0 because there are no numbers on a standard die that meet this condition.
Therefore, the answer is B. rolling a number greater than 1 on a number cube.
A class has 9 girls and 12 boys. What is the probability of a teacher
randomly calling on a boy?
A. 3/4
B. 4/7
C. 3/7
D. 4/3
There are a total of 9 + 12 = 21 students in the class, with 12 of them being boys.
The probability of the teacher randomly calling on a boy is equal to the number of boys divided by the total number of students in the class:
P(calling on a boy) = Number of boys ÷ Total number of students = 12 ÷ 21
We can simplify this fraction by dividing both the numerator and denominator by 3:
P(calling on a boy) = (12 ÷ 3) ÷ (21 ÷ 3) = 4 ÷ 7
Therefore, the probability of the teacher randomly calling on a boy is 4/7.
The answer is B. 4/7.
A jar contains 5 green erasers, 2 striped erasers, 4 pink erasers, and 3 white erasers. You select an eraser at random. What is P(not pink)?
A. 4/14
B. 2/5
C. 5/7
D. 1/7
The total number of erasers in the jar is:
5 + 2 + 4 + 3 = 14
The number of erasers that are not pink is:
5 green erasers + 2 striped erasers + 3 white erasers = 10 erasers
So the probability of selecting an eraser that is not pink is:
P(not pink) = Number of erasers that are not pink ÷ Total number of erasers = 10 ÷ 14
This fraction can be simplified by dividing both the numerator and the denominator by 2:
P(not pink) = 5 ÷ 7
Therefore, the answer is C. 5/7.
On the spinner, what is the probability of spinning a number greater than 2?
A. 1/2
B. 5/8
C. 3/8
D. 1/4
We do not have the spinner or its number of sections mentioned here. Assuming a standard spinner with equal sections numbered 1 through 6, there are four numbers (3, 4, 5, and 6) that are greater than 2 out of a total of six possible outcomes.
Therefore, the probability of spinning a number greater than 2 would be:
Number of outcomes that are greater than 2 ÷ Total number of possible outcomes
= 4 ÷
You write each letter of the word MATHEMATICIAN on a piece of paper and put them in a bag. You randomly select one piece from the bag. What is P(T or A)?
A. 5/13
B. 5/8
C. 1/3
D. 8/13
Hmm idk
Helloo!? AL bot?
Solve the following problems to figure out the 4 digit lock combination
Puzzle 1
2 cats each have
4 kittens. How many kittens do they have altogether?
Ist Digit
You bought one
notebook for $5 and some pens for $3
2x = 48
each. You spent a
3x+2 = 11
total of $26. How many pens did you
Find x
buy
2nd Digit
3rd Digit
Find x
4th Digit
Puzzle 1 Combination *
Your answe—-