jerry mixes the letters F, P, M, M, M, F, P, and j throughly. Without looking., Amanda draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that J will NOT be the letter Amanda selects
There are a total of 8 letters. The letter J appears only once. So, the probability of Amanda selecting J is 1/8. To find the probability of Amanda NOT selecting J, we subtract this probability from 1:
1 - 1/8 = 7/8
As a decimal, this is 0.875, and as a percentage, it is 87.5%. Therefore, the probability that J will NOT be the letter Amanda selects is 7/8, 0.875, or 87.5%.
Christopher mixes the letters S, E, L, E, C, D, E and D thoroughly . Without looking, Ellen draws one letter. Expressed as a fraction, decimal, and percentage. What is the probability that Ellen will not select a consonant
There are a total of 8 letters. To find the probability of Ellen not selecting a consonant, we need to count how many vowels are in the group of letters:
- S is a consonant
- E is a vowel
- L is a consonant
- E is a vowel
- C is a consonant
- D is a consonant
- E is a vowel
- D is a consonant
Out of the 8 letters, 3 are vowels. So, the probability of Ellen selecting a vowel is 3/8. To find the probability of Ellen not selecting a consonant, we can simply add the probabilities of selecting the vowels:
3/8 = 0.375 = 37.5%
Therefore, the probability that Ellen will not select a consonant is 3/8, 0.375, or 37.5%.
Suppose you have a drawer full of white, black, and yellow pairs of socks. If the probability of picking a white pair of socks is 4/9, and the probability of picking black pair of socks is 7/18, what is the probability of picking a yellow pair of socks?
Let's first assign variables for the probability of picking a yellow pair of socks. Let y be the probability of picking a yellow pair of socks. We know that the sum of the probabilities of picking a white, black, or yellow pair of socks is 1. So, we can write an equation:
4/9 + 7/18 + y = 1
To solve for y, we can simplify the equation by finding a common denominator for the first two terms:
8/18 + 7/18 + y = 1
Combining the first two terms, we get:
15/18 + y = 1
Subtracting 15/18 from both sides:
y = 3/18
Simplifying the fraction, we get:
y = 1/6
So, the probability of picking a yellow pair of socks is 1/6.
While shopping, you purchase, four pairs of jeans, six T-shirts, and three 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
13
22
60
72
To find the number of different outfits that can be made using one pair of jeans, one shirt, and one pair of shoes, we need to multiply the number of options for each item.
For jeans, there are 4 options. For T-shirts, there are 6 options. And for shoes, there are 3 options.
So, the total number of different outfits that can be made is:
4 x 6 x 3 = 72
Therefore, the answer is 72. Option (D).
There are four marbles in a big bag with the colors, red, white, blue and green. John pulls out one marble and tossed a coin. How many elements are there in the sample space?
2
4
8
16
When John pulls out one marble, there are 4 possible outcomes: red, white, blue, or green.
When he tosses a coin, there are 2 possible outcomes: heads or tails.
To find the total number of elements in the sample space, we need to multiply the number of outcomes for each action:
4 x 2 = 8
So, there are 8 elements in the sample space. Therefore, the answer is (C) 8.
Allis ice cream shop offers five flavors and 10 toppings. Jails ice cream shop offer seven flavors and seven toppings. Fernandez ice cream shop offers nine flavors and six toppings. Laura’s ice cream shop office 13 flavors and for toppings. If you want flavor of ice cream and one topping, which shop gives you the most options
To find out which shop gives the most options for one flavor of ice cream and one topping, we need to add the number of options for each shop.
- Allis: 5 flavors x 10 toppings = 50 options
- Jails: 7 flavors x 7 toppings = 49 options
- Fernandez: 9 flavors x 6 toppings = 54 options
- Laura's: 13 flavors x 4 toppings = 52 options
Therefore, the ice cream shop that offers the most options for one flavor of ice cream and one topping is Fernadez, with 54 options.
Suppose the probability that it rains in the next two days is 1/3 over tomorrow and 1/6 for the day after tomorrow. What is P(rain tomorrow, then rain the day after)?
To find the probability that it rains tomorrow and the day after tomorrow, we need to multiply the probability of rain tomorrow by the probability of rain the day after tomorrow, given that it rained tomorrow.
P(rain tomorrow, then rain the day after) = P(rain tomorrow) x P(rain the day after | rain tomorrow)
We know that the probability of rain tomorrow is 1/3, so:
P(rain tomorrow, then rain the day after) = (1/3) x P(rain the day after | rain tomorrow)
We also know that if it rains tomorrow, the probability of rain the day after is 1/6, so:
P(rain tomorrow, then rain the day after) = (1/3) x (1/6) = 1/18
Therefore, the probability that it rains tomorrow and the day after tomorrow is 1/18.
William will roll the cubes one time. What is the probability that the numbers showing set up after the role will have a sum of nine
To find the probability of rolling two dice and having a sum of nine, we can list all the possible outcomes. There are 36 possible outcomes when rolling two dice, and the only way to get a sum of nine is by rolling a 3 and a 6, a 4 and a 5, or a 5 and a 4.
So, there are 3 ways to get a sum of 9 out of 36 possible outcomes. Therefore, the probability of rolling two dice and getting a sum of 9 is:
3/36 = 1/12 or approximately 0.083
So, the probability of rolling a sum of 9 on one roll is 1/12 or approximately 0.083.
Pick out of these numbers 1819 3/4 8/9
This question is incomplete. Please provide more information or context so I can understand what you are asking for.
William will roll the cubes one time. What is the probability that the numbers showing face up after roll wheel have a sum of nine?
1/18
1/9
3/4
8/9
The possible outcomes when rolling two dice are 36, each can be represented as an ordered pair (n, m), where n and m are numbers between 1 and 6 representing the outcome of each die. There are a total of three ways to get a sum of 9: (3,6), (4,5), (5,4).
However, since William will roll the dice only once, he is looking for the probability that the sum of the two faces is 9 in a single roll.
The number of ways to get a sum of 9 with just one roll of the dice is 4, because there are four different combinations of dice that add up to 9: (3,6), (4,5), (5,4), and (6,3).
Since there are 6 possible outcomes for each roll, the total number of possible outcomes is 6. Therefore, the probability of getting a sum of 9 in one roll is:
4/36 = 1/9
The answer is (B) 1/9.
How many different arrangements can be made with the letters from the word MATH
12
24
106
1210
The word MATH has four distinct letters, so the number of different arrangements that can be made with these letters is:
4! = 4 x 3 x 2 x 1 = 24
Therefore, the answer is (B) 24.
You want to draft a four player tennis team. There are eight players to choose from. How many different teams can you form?
70
64
40
336
The number of ways to choose 4 players out of 8 without regard to order is:
8 choose 4 = (8!)/(4!*(8-4)!) = (8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 70
Therefore, there are 70 different teams of 4 players that can be formed from 8 players.
The answer is (A) 70.
Write the number of permutations in factorial form. Then simplify. How many different ways can you and five friends sit at your assigned seats when you go to a concert?
6!;120
6!;720
5!;120
5!;30
The number of permutations can be expressed in factorial form as:
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
However, we need to divide by the number of ways that the 5 friends can be arranged within their group of seats, which is given by 5!, or:
5! = 5 x 4 x 3 x 2 x 1 = 120
So, the number of different ways that you and your five friends can sit at your assigned seats is:
6! / 5! = 720 / 120 = 6
Therefore, there are 6 different ways that you and your friends can sit at your assigned seats.
What’s it the answer?
The answer is 6.
Below are the results of tossing a number cube seven times. Find the experimental probability of tossing an even number 6432533.
3/7
2/7
1/2
4/7
The number cube has six possible outcomes, and three of them are even: 2, 4, and 6.
Out of the seven times the cube was tossed, it landed on an even number three times: first, fourth, and seventh toss.
Therefore, the experimental probability of tossing an even number is:
3 (number of even results) / 7 (total number of tosses) = 3/7
So, the answer is (A) 3/7.
How many different ways can a coach select the captain and cocaptain of a team from a group of 20 people?
The number of ways a coach can select the captain and co-captain of a team from a group of 20 people is a combination of choosing 2 people out of 20 without regard to order. The formula for the number of combinations is:
n choose r = (n!)/[(r!)(n-r)!]
In this case, n = 20 and r = 2. Substituting these values into the formula, we get:
20 choose 2 = (20!)/[(2!)(20-2)!] = (20 x 19)/2 = 190
Therefore, there are 190 different ways that a coach can select the captain and co-captain of a team from a group of 20 people.
Pick out of these numbers 40 106 380 420
It is not possible to determine what number to pick based on just the numbers themselves. Please provide more information or context so I can understand what you are asking for.
When building a house, you have the choice of seven different floorplans to different exteriors and two different types of flooring. How many choices are there for one house
To find out how many different choices there are for one house, we need to multiply the number of choices for each option.
- Number of floorplans: 7
- Number of exteriors: 4 (since there are different exteriors for each floorplan)
- Number of flooring types: 2
So, the total number of different choices for one house is:
7 x 4 x 2 = 56
Therefore, there are 56 different choices for one house.
When building a house, you have the choice of seven different floor plans to different exteriors in two different types of flooring how many choices are there for one house?
7
28
4
14
To find the number of choices for one house, we need to multiply the number of choices for each option:
- Number of floor plans: 7
- Number of exteriors for each floor plan: 2
- Number of types of flooring: 2
So, the total number of choices for one house is:
7 x 2 x 2 = 28
Therefore, there are 28 different choices for one house.
The answer is (B) 28.
There are 25 people competitive in a race in how many ways can they finished in first and second place
49
400
600
625
The number of ways that 2 people can be selected from a group of 25 without regard to order is:
25 choose 2 = (25!)/(2!(25-2)!) = (25 x 24)/(2 x 1) = 300
Therefore, there are 300 ways that the first and second place can be finished.
The answer is not among the given options.
There are 25 people competing in a race. In how many ways can they finish in first and 2nd place
The number of ways that 2 people can be selected from a group of 25 without regard to order is:
25 choose 2 = (25!)/(2!(25-2)!) = (25 x 24)/(2 x 1) = 300
Therefore, there are 300 ways that the first and second place can be finished.
So, the answer is 300.
A bag contains tiles with the letters COMBINATIONS. Lee chooses a tile without looking and desert replace it. He chooses a second towels out looking. What is the probability that he will choose the letter O both times.
The probability of selecting the letter O on the first draw is 2/12 because there are two O's in the word COMBINATIONS and there are 12 total letters. After replacing the first tile, the probability of selecting an O on the second draw is also 2/12.
To find the probability of both events happening (selecting an O on the first draw and an O on the second draw), we need to multiply the probabilities:
(2/12) x (2/12) = 1/36
So, the probability that Lee will choose the letter O both times is 1/36.
From my last question pick a number
1/132
1/72
1/66
1/23
From your last question, the probability that Lee will choose the letter O both times is 1/36.
However, none of the options in your follow-up question matches this answer. Please provide more context or clarify your question so I can assist you better.
Haylee is trying to win the grand prize on a game show. Should she try her luck by spinning a wheel with six equal sections labeled from one to six and hope she gets a five, or should she roll two number cubes and hope she gets the same number on both cubes? Explain.
The probability of winning the grand prize by spinning the wheel and landing on a 5 is 1/6. This is because there is only one section labeled with 5 among the six equal sections.
The probability of winning the grand prize by rolling two number cubes and getting the same number on both cubes depends on the number of possible outcomes. When rolling two dice, there are 36 possible outcomes, each represented as an ordered pair (n, m), where n and m are numbers between 1 and 6 representing the outcome of each die. Out of these 36 possible outcomes, there are 6 outcomes where the two numbers are the same: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). So, the probability of rolling the same number on both dice is:
6 (number of favorable outcomes) / 36 (total number of outcomes) = 1/6
Therefore, the probability of winning the grand prize by rolling two number cubes and getting the same number on both cubes is also 1/6.
So, both options have the same probability of winning the grand prize, which is 1/6. Therefore, it doesn't matter whether Haylee spins the wheel or rolls two number cubes as the probability of winning is the same.