A word consists of 9 letters in total, of which 5 are consonants and 4 are vowels. Three letters are chosen at random without replacement. What is the probability that more than one vowel (that is, at least two vowels) will be selected?

Introduction / Context:
This question checks understanding of selections without replacement and how to calculate probabilities involving phrases like more than one vowel. It requires counting favorable combinations of letters in a situation where the composition of the set is known. The student must carefully interpret more than one vowel as meaning at least two vowels and use basic combinatorics to count these cases.

Given Data / Assumptions:

• Total letters in the word = 9.
• Consonants = 5 and vowels = 4.
• Three letters are selected at random without replacement.
• Each three letter selection is equally likely.
• More than one vowel means exactly two vowels or exactly three vowels.

Concept / Approach:
The total number of ways to choose 3 letters from 9 is computed using combinations. Favourable outcomes occur in two disjoint cases: exactly two vowels and one consonant, or exactly three vowels. We calculate the number of combinations in each case, add them, and divide by the total number of combinations. This is a standard example of using combination formulas in probability.

Step-by-Step Solution:
Step 1: Compute total number of ways to choose 3 letters from 9 letters: C(9,3) = 84.Step 2: For exactly two vowels and one consonant, choose 2 vowels from 4 and 1 consonant from 5. That gives C(4,2) * C(5,1) = 6 * 5 = 30 ways.Step 3: For exactly three vowels, choose 3 vowels from 4. That gives C(4,3) = 4 ways.Step 4: Total favorable selections for more than one vowel is 30 + 4 = 34.Step 5: Probability = favorable outcomes / total outcomes = 34 / 84.Step 6: Simplify 34 / 84 by dividing numerator and denominator by 2 to get 17 / 42.
Verification / Alternative check:
To verify, we may compute the probabilities of the complementary events (zero vowels and exactly one vowel) and check that all probabilities add to 1. Zero vowels means all three letters are consonants, which has C(5,3) = 10 ways. Exactly one vowel gives C(4,1) * C(5,2) = 4 * 10 = 40 ways. Adding favorable cases for more than one vowel (34) plus 10 and 40 gives a total of 84, which matches C(9,3), confirming that our counting is consistent.

Why Other Options Are Wrong:
13/42 would correspond to a smaller number of favorable outcomes than we actually count and does not match the combination counts. 5/42 is even smaller and clearly incorrect after checking all cases. 11/42 is not supported by any natural breakdown of cases. 1/2 would suggest 42 favorable outcomes out of 84, which contradicts the exact count of 34 favorable selections.

Common Pitfalls:
Students sometimes misinterpret more than one vowel as exactly two vowels and forget the three vowel case. Another frequent mistake is to use permutations instead of combinations, which would overcount arrangements that differ only in order. It is also easy to miscount by accidentally using 9 choose 2 or other incorrect totals. Careful listing of cases and use of combination formulas helps avoid these errors.

Final Answer:
The probability that more than one vowel will be selected is 17/42.