Question

Let \( S \) be the set of all the words that can be formed by arranging all the letters of the word GARDEN. From the set \( S \), one word is selected at random. The probability that the selected word will NOT have vowels in alphabetical order is:

• A. \( \frac{1}{4} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{2} \)

Hint:

For problems involving probability, it is often useful to calculate the complementary event and subtract from 1.

Answer 1

The Correct Option is D

We are given the word GARDEN, which consists of the following letters: G, A, R, D, E, N. Among these letters, the vowels are A and E. To find the probability that the selected word will NOT have vowels in alphabetical order, we proceed as follows: 
Step 1: Total number of arrangements. 
Since there are 6 distinct letters in the word GARDEN, the total number of ways to arrange these letters is: \[ {Total arrangements} = 6! = 720 \] 
Step 2: Number of favorable cases (vowels in alphabetical order). 
For the vowels A and E to be in alphabetical order, the positions of A and E must be such that A appears before E. The total number of ways to arrange the 6 letters such that A appears before E is: \[ {Favorable cases} = \binom{6}{2} \cdot 4! = 15 \cdot 24 = 360 \] 
Step 3: Probability calculation. 
The probability that the selected word will have vowels in alphabetical order is: \[ P = \frac{360}{720} = \frac{1}{2} \] Therefore, the probability that the selected word will NOT have vowels in alphabetical order is: \[ P({Not in order}) = 1 - \frac{1}{2} = \frac{1}{2} \]