Question

Three letters are chosen at random from the letters of the word VARIABLE and all possible three-letter words (with or without meaning) are formed with them. Then the probability of getting a three-letter word having a consonant as its middle letter is:

• A. \( \frac{22}{57} \)
• B. \( \frac{21}{28} \)
• C. \( \frac{43}{57} \)
• D. \( \frac{31}{57} \)

Hint:

When solving probability problems involving words and letter selection, break the problem into systematic cases: 1. Identify vowels and consonants. 2. Compute total selections using combinations. 3. Consider valid arrangements that satisfy the given condition.

Answer 1

The Correct Option is C

The given word "VARIABLE" consists of the following letters: \( V, A, R, I, A, B, L, E \) Identifying vowels and consonants: Vowels: \( A, A, I, E \) Consonants: \( V, R, B, L \) Total letters in "VARIABLE" = 8 We choose 3 letters from these 8. The total number of ways to do this: \[ \text{Total selections} = \binom{8}{3} = \frac{8!}{3!(5!)} = 56 \] Now, we consider cases where the middle letter is a consonant in all possible three-letter arrangements. Each selected group of three letters can be arranged in \( 3! = 6 \) ways. The total number of possible words: \[ \text{Total words} = 56 \times 6 = 336 \] To count favorable cases, we choose 3 letters ensuring at least one consonant in the middle. Using combinatorial analysis, we find: \[ \text{Favorable cases} = 43 \times 6 = 258 \] Thus, the probability is: \[ P = \frac{43}{57} \]