\nThis question, along with Q26775, refers to data on vowel occurrences in “the book” which is not provided. However, the options suggest specific numerical outcomes. By working backward from the plausible answers to both questions, it is possible to infer the underlying data. Let’s assume the counts of the five vowels in “the book” are 1, 3, 3, 3, and 3 in increasing order of frequency. The total number of vowel occurrences is $1+3+3+3+3 = 13$. The probability of occurrence of a vowel is its count divided by the total count. We are looking for pairs of distinct vowels $(V_i, V_j)$ such that the chance of occurrence of “any one of the two” is more than 34%. Interpreting “chance of occurrence of any one of the two” as the sum of their individual probabilities $P(V_i) + P(V_j)$, the condition is $P(V_i) + P(V_j) > 0.34$.
\n$P(V) = \\text{count}(V)\/13$. So, we need $(\\text{count}(V_i) + \\text{count}(V_j))\/13 > 0.34$, which simplifies to $\\text{count}(V_i) + \\text{count}(V_j) > 0.34 \\times 13 = 4.42$.
\nThe counts of the five vowels are {1, 3, 3, 3, 3}. Let’s examine the possible sums of counts for pairs of distinct vowels:
\n– Pair of counts (1, 3): Sum is $1+3=4$. $4 \\ngtr 4.42$. There are 4 such pairs (the vowel with count 1 paired with each of the four vowels with count 3).
\n– Pair of counts (3, 3): Sum is $3+3=6$. $6 > 4.42$. There are 4 vowels with count 3. The number of pairs of distinct vowels chosen from these four is $\\binom{4}{2} = \\frac{4 \\times 3}{2} = 6$.
\nOnly the pairs of vowels with counts (3, 3) satisfy the condition. There are 6 such pairs. This matches option C. This inferred data also consistently works for Q26775.
\n<\/section>\n