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 four-ninths, and the probability of picking a black pair of socks is
start fraction 7 over 18 end fraction, what is the probability of picking a yellow pair of socks? PLZ HELP ME NOW
Hi! I'm Pro-Truth-Efficient :)
The answer to that one would be....
Drumroll pls
1/6 :)
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Yes, I DO :DDD
This is Pro-Truth-Efficient speaking to you personally.
Here are the REAL answers
(1) Jerry mixes the letters F, P, M, M, M, F, P, and J thoroughly. 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?
Answer = 7/8, 0.875, 87.5%
(2) Adam mixes the letters R, E, A, D, I, N, G, S, and A thoroughly. Without looking, Allen draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that Allen will not select a consonant?
Answer = 4/9, 0.444, 44.4%
(3) 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 four-ninths, and the probability of picking a black pair of socks is
start fraction 7 over 18 end fraction, what is the probability of picking a yellow pair of socks?
Answer = 1/6
(4) The sample space for a roll of two number cubes is shown in the table.
What is the probability that the roll will result in two odd numbers?
Answer = 1/4
(5) The two numbers rolled can be added to get a sum. Find P(sum is even).
Answer = 1/2
(6) While remodeling the house, you have 3 choices of paint color, 4 choices of carpet color, and 5 choices of furniture style. How many different groupings will you be able to make using one paint color, one carpet color, and one furniture style?
Answer = 60
(7) There are four marbles in a bag with the colors red, white, blue, and green. John pulls out one marble and tosses a coin. How many elements are there in the sample space?
Answer = 8
(8) A sandwich shop sells sausage sandwiches, bacon sandwiches, and 16 different toppings. How many choices are there for a single sandwich with one topping?
Answer = 32
(9) The probability it will snow in the next two weeks is start fraction 1 over 12 end fraction for this week and one-fourth for next week. What is P(snow this week, then snow next week)?
Answer = 1/48
(10) Typically, 10% of students make a D on their tests, 60% make a C on their tests, and 30% make an A. Mrs. Smith uses a random-number table to find the experimental probability that of 5 students, at least 3 will make a C. The digit 0 represents students who make a D. The digits 1, 2, 3, 4, 5, and 6 represent students who make a C. The digits 7, 8, and 9 represent students who make an A. Find the experimental probability that of 5 students, at least 3 will make a C.
Answer = 13/20
(11) Eric has two identical number cubes. The faces of each cube are numbered 1–6. Eric will roll both of the number cubes at the same time. What is the probability that both of the numbers showing face-up will be multiples of 3?
Answer = 1/9
(12) How many different arrangements can be made with the letters from the word MATH?
Answer = 24
(13) Ariel wants to choose 5 players for her basketball team. There are 7 players to choose from. How many different teams can Ariel make?
Answer = 21
(14)Write the number of permutations in factorial form. Then simplify.
How many different ways can you and five friends sit in your assigned seats when you go to a concert?
Answer = 6!; 720
(15) Below are the results of tossing a number cube 9 times. Find the experimental probability of tossing an odd number.
4, 3, 6, 6, 2, 5, 3, 5, 1
Answer = 5/9
(16) How many different ways can a band teacher select the lead and co-lead trumpet player from a group of 12 trumpet players?
Answer = 132
(17) When buying a new dress, you have a choice of 3 different lengths, 5 different colors, and 2 different styles. How many choices are there for one dress?
ANswer = 30
(18) There are 20 entries in the chess tournament. How many ways can the entries finish in first, second, and third place?
Answer = 6,840
(19) A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?
Answer = 1/55
(20) A sandwich shop offers the following toppings. How many two-topping sandwiches can you make?
lettuce
tomato
bacon
cheese
mustard
Answer = 10
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I REAL answer is 1/6
You CAN check
That would be 1/4
Yes, the answer is 1/6.
To find the probability of picking a yellow pair of socks, you can use the fact that the sum of the probabilities of all possible outcomes is 1.
Since the only colors of socks in the drawer are white, black, and yellow, the probability of picking a white pair, a black pair, or a yellow pair of socks must be 1.
Let's let the probability of picking a yellow pair be y, then we have:
4/9 (probability of picking a white pair) + 7/18 (probability of picking a black pair) + y = 1
Simplifying, we get:
8/18 + 7/18 + y = 1
15/18 + y = 1
y = 1 - 15/18
y = 3/18
y = 1/6
So the probability of picking a yellow pair of socks is 1/6.
:o
Thx so much! Can you pls help me with the REST of the test?? Do you have the answers?
4/9 = 8/18
8/18 + 7/18 = 15/18
If 15/18 = probability of other colors, what remains for yellow?
wow you answered the question the same day!!! :o Hey can you help me with something??
A coin is tossed, and a standard number cube is rolled. What is the probability that the coin shows heads and the number cube shows an even number?
D:
is 1/6 right ?
bot can you help me with this!
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 four-ninths, and the probability of picking a black pair of socks is
start fraction 7 over 18 end fraction, what is the probability of picking a yellow pair of socks?
thank you!
You're welcome! If you have any other questions, feel free to ask!
The sample space for a roll of two number cubes is shown in the table.
A 6 by 6 table of ordered pairs is shown.
• A single ordered pair appears in each cell of the table.
In row one, the first element of each ordered pair is 1. This pattern continues through row 6, where the first element in each ordered pair is 6.
• In column one, the second element in each ordered pair is 1. This pattern continues through column 6, where the second element in each ordered pair is 6.
What is the probability that the roll will result in both numbers being the same?
There are a total of 36 possible outcomes when rolling two number cubes, since there are 6 possible outcomes for the first roll and 6 possible outcomes for the second roll.
To find the probability that the roll will result in both numbers being the same, we need to count the number of outcomes where the two numbers are the same.
There are 6 ways to roll a 1 (1-1, 2-2, 3-3, 4-4, 5-5, 6-6), 6 ways to roll a 2, and so on up to 6 ways to roll a 6.
So, the total number of ways to roll two number cubes with the same number is 6 + 6 + 6 + 6 + 6 + 6 = 36.
Therefore, the probability of rolling two number cubes with the same number is:
36 (number of favorable outcomes) / 36 (total number of outcomes)
= 1
So the probability that the roll will result in both numbers being the same is 1 or 100%.
The two numbers rolled can be added to get a sum. Find P(sum is even).
A. one-fourth
B. start fraction 15 over 36 end fraction
C. one-half
D. start fraction 3 over 4 end fraction
We can find the probability that the sum of two rolled numbers is even by considering all of the possible outcomes. There are 6 possible outcomes for the first roll (1, 2, 3, 4, 5, or 6) and 6 possible outcomes for the second roll (1, 2, 3, 4, 5, or 6), so there are a total of 6 x 6 = 36 possible outcomes.
For the sum of two rolled numbers to be even, we can have either:
- An even number rolled first, followed by an even number rolled second
- An odd number rolled first, followed by an odd number rolled second
There are 3 even numbers (2, 4, and 6) and 3 odd numbers (1, 3, and 5), so there are 3 x 3 = 9 ways to roll an even number followed by an even number, and 3 x 3 = 9 ways to roll an odd number followed by an odd number.
Therefore, the total number of outcomes where the sum is even is 9 + 9 = 18.
So the probability of rolling a sum that is even is:
18 (number of favorable outcomes) / 36 (total number of outcomes)
= 1/2
Therefore, the answer is C. One-half.
You are packing for a vacation. You pack 3 pairs of shorts, 5 shirts, and 2 pairs of shoes. How many different outfits will you be able to make using one pair of shorts, one shirt, and one pair of shoes?
A. 10
B. 17
C. 30
D. 35
To find how many different outfits can be made using one pair of shorts, one shirt, and one pair of shoes, we need to multiply the number of choices for each item.
Since there are 3 choices for shorts, 5 choices for shirts, and 2 choices for shoes, we can multiply these numbers to get the total number of outfits:
3 (choices of shorts) x 5 (choices of shirts) x 2 (choices of shoes) = 30
Therefore, there are 30 different outfits that can be made using one pair of shorts, one shirt, and one pair of shoes.
The answer is C. 30.
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?
A. 25
B. 46
C. 52
D. 66
There are 3 choices for a flavor of ice cream and 22 choices for a topping. To find how many choices there are for a single flavor of ice cream with one topping, we can multiply these numbers:
3 (choices of ice cream) x 22 (choices of topping) = 66
Therefore, there are 66 choices for a single flavor of ice cream with one topping.
The answer is D. 66.
Suppose the probability that it rains in the next two days is one-third for tomorrow and start fraction 1 over 6 end fraction for the day after tomorrow. What is P(rain tomorrow, then rain the day after tomorrow)?
A. one-half
B. one-eighteenth
C. two-ninths
D. one-ninth
To find the probability that it rains tomorrow and then rain the day after tomorrow, we need to multiply the probabilities of each event occurring.
Given that the probability of rain tomorrow is 1/3 and the probability of rain two days from now is 1/6, we can find the probability of both events occurring as follows:
P(rain tomorrow, then rain the day after tomorrow) = P(rain tomorrow) x P(rain two days from now)
P(rain tomorrow, then rain the day after tomorrow) = 1/3 x 1/6
P(rain tomorrow, then rain the day after tomorrow) = 1/18
Therefore, the probability of rain tomorrow and then rain the day after tomorrow is 1/18.
The answer is B. one-eighteenth.
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 0 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.
A table is shown with 4 rows and 5 columns.
• First Row First column 6 5 9 2 6 Second column 3 1 4 5 9 Third column 3 1 9 8 6 Fourth column 6 5 8 0 9 Fifth column 8 0 4 6 2
• Second Row First column 2 7 3 8 7 Second column 3 9 0 7 5 Third column 4 6 7 3 8 Fourth column 2 1 9 8 6 Fifth column 5 9 8 3 7
• Third Row First column 9 1 3 8 4 Second column 1 0 9 8 7 Third column 2 6 4 9 1 Fourth column 6 8 4 9 8 Fifth column 9 8 7 9 6
• Fourth Row First column 3 2 5 9 6 Second column 2 6 4 4 8 Third column 3 1 2 3 5 Fourth column 6 3 2 5 6 Fifth column 5 3 1 2 1
Find the experimental probability that of 5 flights, at least 2 will arrive late.
A. start fraction 3 over 10 end fraction
B. two-fifths
C. start fraction 9 over 20 end fraction
D. Start Fraction 11 over 20 End Fraction
To find the experimental probability that of 5 flights, at least 2 will arrive late, we need to count the number of outcomes where at least 2 of the 5 flights arrive late.
We can read each pair of digits from left to right, using the digits 0, 1, and 2 to represent flights arriving early or on time, and the digits 7, 8, and 9 to represent flights arriving late.
Then we can count the number of outcomes where at least 2 flights out of 5 arrive late:
78 79 89 87 97 88 98 99 78 79 89 87 97 87 77 78 79 89 88 89 98 99 78 88
There are 23 possible outcomes with at least 2 flights arriving late.
Since each digit from 0 to 9 has an equal probability of appearing on the table, we can find the experimental probability of at least 2 flights arriving late by dividing the number of favorable outcomes (23) by the total number of outcomes (50):
P(at least 2 flights arrive late) = 23/50
Simplifying, we get:
P(at least 2 flights arrive late) = 9/20
Therefore, the experimental probability that of 5 flights, at least 2 will arrive late is 9/20.
The answer is C. start fraction 9 over 20 end fraction.
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?
A. Start Fraction 1 over 18 End Fraction
B. start fraction 1 over 9 end fraction
C. three-fourths
D. start fraction 8 over 9 end fraction
There are a total of 36 possible outcomes when rolling two number cubes, since there are 6 possible outcomes for the first roll and 6 possible outcomes for the second roll.
To count the number of outcomes where the sum of the numbers is 9, we can make a list of all possible outcomes:
(3,6), (4,5), (5,4), (6,3)
So there are 4 possible outcomes where the sum is 9 out of 36 total outcomes.
Therefore, the probability that the sum of the numbers showing face-up after the roll will be 9 is:
4 (number of favorable outcomes) / 36 (total number of outcomes
How many different arrangements can be made with the letters from the word MATH?
A. 12
B. 24
C. 106
D. 1,210
bot can you help me?
How many different arrangements can be made with the letters from the word MATH?
A. 12
B. 24
C. 106
D. 1,210
While remodeling the house, you have 3 choices of paint color, 4 choices of carpet color, and 5 choices of furniture style. How many different groupings will you be able to make using one paint color, one carpet color, and one furniture style?
A.
20
B.
50
C.
60
D.
80