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?
A. seven-eighths, 0.875, 87.5%
B. The fraction is 5 over 8., 0.625, 62.5%
C. one-eighth, 0.125, 12.5%
D. The term shows 8 over 7., 1.14, 11.4%
There are a total of 8 letters, and only 1 of them is a J. Therefore, the probability that Amanda selects J is $\dfrac{1}{8}$. The probability that she does not select J is $1-\dfrac{1}{8}=\dfrac{7}{8}$.
Converting to decimal and percentage form:
$\frac{7}{8}=0.875=87.5\%$
The answer is $\boxed{\textbf{(A)}\ \frac{7}{8}, 0.875, 87.5\%}$.
Tammy mixes the letters S, C, H, O, O, and L 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 5 over 6 End Fraction, 0.833, 83.3%
B. Start Fraction 2 over 3 End Fraction, 0.667, 66.7%
C. one third, 0.333, 33.3%
D. Start Fraction 3 over 2 End Fraction, 1.5, 15%
There are two vowels in the six-letter set: $\{{\rm S, C, H, O, O, L}\}$: $\{{\rm O, O}\}$ (it does not matter that there are two of them). The probability that Alan chooses a non-vowel on his first try is $\frac{4}{6}=\frac{2}{3}$. Expressed as a decimal and percentage, this probability is $0.\overline{6}=0.667=66.7\%$, so the answer is $\boxed{\textbf{(B)}\ \frac{2}{3}, 0.667, 66.7\%}$
Suppose you have a bag of red, blue, and yellow marbles. If the probability of picking a red marble is start fraction 1 over 6 end fraction and the probability of picking a yellow marble is one-half, what is the probability of picking a blue marble?
A. one-fourth
B. one-third
C. two-thirds
D. one-eighth
We can start by using the fact that the sum of all the probabilities of picking a red, blue, or yellow marble must equal 1. Let the probability of picking a blue marble be $x$. Then:
$\text{Probability of red} + \text{Probability of blue} + \text{Probability of yellow} = 1$
$\frac{1}{6} + x + \frac{1}{2} = 1$
$x = \frac{1}{3}$
Therefore, the probability of picking a blue marble is $\boxed{\textbf{(B)}\ \frac{1}{3}}$.
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 one even and one odd number?
A. one-ninth
B. one-fourth
C. one-third
D. one-half
We can count the number of outcomes that have one even and one odd number by breaking it up into cases:
Case 1: Even number first
There are 3 even numbers and 3 odd numbers, so there are $3\cdot3=9$ outcomes with an even number first.
Case 2: Odd number first
Using the same reasoning, there are another 9 outcomes with an odd number first.
Therefore, there are a total of $9+9=18$ outcomes that have one even and one odd number.
Since there are $6\cdot6=36$ possible outcomes in total (each of the 6 numbers on the first cube can be paired with each of the 6 numbers on the second cube), the probability of rolling one even and one odd number is $\dfrac{18}{36}=\boxed{\textbf{(C)}\ \frac{1}{3}}$.
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 less than 4).
A. start fraction 1 over 12 end fraction
B. Start Fraction 5 over 36 End Fraction
C. Start Fraction 2 over 9 End Fraction
D. start fraction 11 over 12 end fraction
The only two ordered pairs with a sum less than 4 are $(1,1)$ and $(1,2)$. Therefore, there are only 2 favorable outcomes out of a total of 36 possible outcomes, since there are 6 choices for the first number and 6 choices for the second number.
Therefore, the probability of rolling a sum less than 4 is $\dfrac{2}{36}=\dfrac{1}{18}$. Converting to fraction and decimal form, the answer is $\boxed{\textbf{(A)}\ \frac{1}{12}, 0.0833}$.
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