There are 8 letters in total. Out of these, only the letters S, L, T, and D are consonants. Therefore, there are 4 consonants and 4 vowels (E). The probability of Allen selecting a consonant is $\dfrac{4}{8} = \dfrac{1}{2}$ or 0.5 or 50\%. Therefore, the probability of Allen not selecting a consonant is $1 - \dfrac{1}{2} = \dfrac{1}{2}$ or 0.5 or 50\%.
Therefore, the answer is $\boxed{\textbf{(D) } \text{one-half, 0.5, 50\%}}$.
Christopher mixes the letters S, E, L, E, C, T, E, and D 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. three-eighths, 0.375, 37.5%
B. start fraction 3 over 5 end fraction, 0.6, 60%
C. start fraction 5 over 8 end fraction, 0.625, 62.5%
D. one-half, 0.5, 50%.
1 answer