Jacob mixes the letters J, K, L, J, K, M, N, and P thoroughly. Without looking, Terry draws one letter. Expressed as a fraction, decimal, and percentage, what is the probability that K will not be the letter Terry selects?
A. one-fourth, 0.25, 25%
B. three-fourths, 0.75, 75%
C. Start Fraction 4 over 3 End Fraction, 1.33%, 13.3%
D. start fraction 5 over 8 end fraction, 0.625, 62.5%
There are a total of 8 letters, and Terry has an equal chance of drawing any one of them.
There are 2 K's out of 8 letters, so the probability of Terry drawing a K is 2/8 or 1/4.
Therefore, the probability of Terry not drawing a K is 1 - 1/4 = 3/4.
Expressed as a fraction, decimal, and percentage, this is:
$\frac{3}{4}$, 0.75, 75%
The answer is $\boxed{\textbf{(B)}}$.
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?
A. Start Fraction 9 over 5 End Fraction, 1.8, 18%
B. nine-fourths, 2.25, 22.5%
C. four-ninths, 0.444, 44.4%
D. start fraction 5 over 9 end fraction, 0.556, 55.6%
Out of the 9 letters, there are 5 consonants (R, D, N, G, and S) and 4 vowels (E, A, I, and the second A).
Therefore, the probability of Allen selecting a consonant is 5/9.
The probability of Allen not selecting a consonant is the complement of this probability, which is 1 - 5/9 = 4/9.
Expressed as a fraction, decimal, and percentage, this is:
$\frac{4}{9}$, 0.444, 44.4%
The answer is $\boxed{\textbf{(C)}}$.
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?
A. start fraction 1 over 6 end fraction
B. Start Fraction 5 over 12 End Fraction
C. Start Fraction 7 over 15 End Fraction
D. Start Fraction 16 over 27 End Fraction
We know that the sum of the probabilities of picking a white, black, and yellow pair of socks is 1. Therefore, we can use this fact to find the probability of picking a yellow pair of socks:
P(white) + P(black) + P(yellow) = 1
We are given that:
P(white) = 4/9
P(black) = 7/18
Substituting these values into the equation above, we get:
4/9 + 7/18 + P(yellow) = 1
Multiplying both sides by 18, we get:
8 + 7 + 18P(yellow) = 18
Simplifying and solving for P(yellow), we get:
18P(yellow) = 3
P(yellow) = 1/6
Therefore, the probability of picking a yellow pair of socks is:
$\frac{1}{6}$, which is answer choice $\boxed{\textbf{(A)}}$.
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 two odd numbers?
A. one-ninth
B. one-fourth
C. one-third
D. start fraction 4 over 9 end fraction
There are a total of 6 x 6 = 36 possible outcomes when rolling two number cubes.
We want to count the number of outcomes where both numbers are odd. There are 3 odd numbers on each cube (1, 3, and 5), so there are 3 x 3 = 9 outcomes where both numbers are odd.
Therefore, the probability of rolling two odd numbers is 9/36 or 1/4.
Expressed as a fraction, decimal, and percentage, this is:
$\frac{1}{4}$, 0.25, 25%
The answer is $\boxed{\textbf{(B)}}$.
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.
The two numbers rolled can be added to get a sum. Find P(sum is greater than 5).
A. Start Fraction 5 over 6 End Fraction
B. thirteen-eighteenths
C. Start Fraction 5 over 18 End Fraction
D. one-third
To find the probability that the sum is greater than 5, we can count the number of outcomes where this is the case.
If we examine the table, we can see that there are 11 possible outcomes where the sum is greater than 5:
(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3)
There are a total of 6 x 6 = 36 possible outcomes when rolling two number cubes. Therefore, the probability of the sum being greater than 5 is 11/36.
Expressed as a fraction, decimal, and percentage, this is:
$\frac{11}{36}$, 0.306, 30.6%
The answer is $\boxed{\textbf{(B)}}$.
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
You have 3 choices of shorts, 5 choices of shirts, and 2 choices of shoes.
Using the Multiplication Principle, we can determine the total number of outfits by multiplying the number of choices for each category:
Number of outfits = 3 x 5 x 2 = 30
Therefore, you will be able to make 30 different outfits.
The answer is $\boxed{\textbf{(C)}}$.
A spinner has 3 equal sections: red, white, and blue. John spins the spinner and tosses a coin. Which shows the sample space for spinning the spinner and tossing the coin?
A. A table with 3 rows and 4 columns is shown.The second column of the first row reads Red. The third column of the first row reads White. The fourth column of the first row reads Blue. The first column of the second row reads Heads. The second column of the second row reads Heads, Red. The third column of the second row reads Heads, White. The fourth column of the second row reads Heads, Blue. The first column of the third row reads Tails. The second column of the third row reads Heads, Red. The third column of the third row reads Heads, White. The fourth column of the third row reads Heads, Blue.
The first column of the second row reads 1750 dash 1800. The second column of the second row has the number 33. 1800 dash 1850. The second column of the third row has the number 57. The first column of the fourth row reads 1850 dash 1900. The second column of the fourth row has the number 81. The first column of the fifth row reads 1900 dash 1950. The second column of the fifth row has the number 96. The first column of the sixth row reads 1950 dash 2000. The second column of the sixth row has the number 110.
B. A table with 3 rows and 3 columns is shown.The second column of the first row reads Red. The third column of the first row reads White. The first column of the second row reads Heads. The second column of the second row reads Heads, Red. The third column of the second row reads Heads, White. The first column of the third row reads Tails. The second column of the third row reads Heads, Red. The third column of the third row reads Heads, White.
C. A table with 2 rows and 4 columns is shown.The second column of the first row reads Red. The third column of the first row reads White. The fourth column of the first row reads White. The first column of the second row reads Heads. The second column of the second row reads Heads, Red. The third column of the second row reads Heads, White. The fourth column of the second row reads Heads, Blue.
D. A table with 3 rows and 4 columns is shown.The second column of the first row reads Red. The third column of the first row reads White. The fourth column of the first row reads Blue. The first column of the second row reads Heads. The second column of the second row reads Heads, Red. The third column of the second row reads Heads, White. The fourth column of the second row reads Heads, Blue. The first column of the third row reads Tails. The second column of the third row reads Tails, Red. The third column of the third row reads Tails, White. The fourth column of the third row reads Tails, Blue.
The sample space for spinning the spinner and tossing a coin consists of all possible outcomes of these two events.
The spinner has 3 equal sections, so there are 3 equally likely outcomes for spinning the spinner: red, white, or blue. The coin toss has 2 equally likely outcomes: heads or tails. Therefore, the total number of possible outcomes is:
3 x 2 = 6
To represent the sample space in a table, we can list the possible outcomes for each event and combine them in the following way:
| | Red | White | Blue |
|---|---|---|---|
| Heads | Heads, Red | Heads, White | Heads, Blue |
| Tails | Tails, Red | Tails, White | Tails, Blue |
Therefore, the answer is $\boxed{\textbf{(D)}}$.
A sandwich shop sells sausage sandwiches, bacon sandwiches, and 16 different toppings. How many choices are there for a single sandwich with one topping?
A. 18
B. 24
C. 32
D. 34
There are two parts to this sandwich order: choosing the type of sandwich and choosing the topping.
There are 2 types of sandwiches (sausage and bacon) and 16 different toppings.
Using the Multiplication Principle, we can determine the total number of choices by multiplying the number of choices for each category:
Number of choices = 2 x 16 = 32
Therefore, there are 32 different choices for a single sandwich with one topping.
The answer is $\boxed{\textbf{(C)}}$.
The probability that James will win two races in the next two days is one-eighth for tomorrow and one-tenth for the day after tomorrow. What is P(win tomorrow, then win the day after tomorrow)?
A. start fraction 1 over 80 end fraction
B. start fraction 1 over 40 end fraction
C. Start Fraction 1 over 18 End Fraction
D. Start Fraction 2 over 9 End Fraction
If the probability of James winning a race tomorrow is $\frac{1}{8}$ and the probability of him winning the race the day after tomorrow is $\frac{1}{10}$, then the probability of him winning both races is the product of the two probabilities:
$P(win~tomorrow~and~win~day~after~tomorrow) = \frac{1}{8} \times \frac{1}{10} = \frac{1}{80}$
Expressed as a fraction, this is $\frac{1}{80}$, which is answer choice $\boxed{\textbf{(A)}}$.
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.
74510 62548 71205 07819 02478
15678 02389 01258 23789 34052
84150 27549 05689 07821 56892
16059 07894 12356 01548 23689
Find the experimental probability that of 5 students, at least 3 will make a C.
A. Start Fraction 7 over 20 End Fraction
B. Start Fraction 9 over 20 End Fraction
C. Start Fraction 11 over 20 End Fraction
D. Start Fraction 13 over 20 End Fraction
We need to count the number of outcomes in which at least 3 students make a C. We can do this by listing all the outcomes in which 3 or more students make a C and then counting the total number of such outcomes.
From the table, we can see that the following outcomes have at least 3 students making a C:
- 74510
- 62548
- 71205
- 23789
- 34052
- 84150
- 27549
- 07821
- 56892
- 07894
- 12356
- 01548
- 23689
There are a total of 20 outcomes, so the experimental probability of at least 3 students making a C is:
$\frac{13}{20}$
Expressed as a fraction and percentage, this is:
$\frac{13}{20}$, 0.65, 65%
The answer is $\boxed{\textbf{(D)}}$.
Elizabeth has two identical number cubes. Both cubes have faces numbered 1 through 6. If Elizabeth rolls each cube once, what is the probability that the sum of the two numbers on the top faces will be 10?
A. start fraction 1 over 36 end fraction
B. start fraction 1 over 12 end fraction
C. one-tenth
D. start fraction 1 over 9 end fraction
There are a total of 6 x 6 = 36 possible outcomes when rolling two number cubes.
To count the number of outcomes where the sum of the top faces is 10, we can use the following table:
| | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
We can see from the table that there is only 1 outcome where the sum of the top faces is 10, namely (4, 6). Therefore, the probability of rolling a sum of 10 is:
$\frac{1}{36}$
Expressed as a fraction, decimal, and percentage, this is:
$\frac{1}{36}$, 0.028, 2.8%
The answer is $\boxed{\textbf{(A)}}$.
How many different arrangements can be made with the letters from the word MATH?
A. 12
B. 24
C. 106
D. 1,210
There are 4 letters in the word MATH, so there are 4 choices for the first letter, 3 choices for the second letter (since one letter has already been used), 2 choices for the third letter, and 1 choice for the fourth letter.
Using the Multiplication Principle, we can determine the total number of arrangements by multiplying the number of choices for each letter:
Number of arrangements = 4 x 3 x 2 x 1 = 24
Therefore, there are 24 different arrangements that can be made with the letters from the word MATH.
The answer is $\boxed{\textbf{(B)}}$.
Jason wants to choose 9 players for his track team. There are 12 players to choose from. How many different teams can Jason make?
A. 21
B. 112
C. 220
D. 306