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

Step 1: Total Number of Arrangements

The given word "GARDEN" contains 6 distinct letters: G, A, R, D, E, N.
The total number of possible arrangements of these 6 letters is: \[ \text{Total arrangements} = 6! = 720 \]

Step 2: Number of Favorable Cases (Vowels in Alphabetical Order)

The vowels in the word are A and E.
For the vowels to appear in alphabetical order (A before E), the number of valid arrangements is: \[ \binom{6}{2} \cdot 4! = 15 \cdot 24 = 360 \]

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: \[ P(\text{Not in order}) = 1 - \frac{1}{2} = \frac{1}{2} \]

Final Answer: \( \frac{1}{2} \)