Directions: The next three items are based on a survey on occurrence of vowels in a certain book irrespective of whether they are in upper or lower case. Vowel: A | E | I | O | U Percentage: 20 | 45 | 15 | 8 | 12 For how many pairs of vowels is the chance of occurrence of any one of the two more than 34% in the book?

Correct — B, 5. This is the first of three questions built on one printed table — A 20, E 45, I 15, O 8, U 12. The chance that a randomly chosen vowel is one of two given vowels is simply the sum of their two percentages, because one occurrence cannot be two vowels at once. Five vowels give ten possible pairs, so write out all ten sums: A+E = 65, A+I = 35, A+O = 28, A+U = 32, E+I = 60, E+O = 53, E+U = 57, I+O = 23, I+U = 27, O+U = 20. Exactly five of them exceed 34% — A+E, A+I, E+I, E+O and E+U. The pattern is worth noticing: every pair containing E clears the bar on its own because E alone is 45, and only one pair without E, namely A+I at 35, squeaks over.

• (a)4 — This is the count of pairs containing E. It misses A+I at 35%, the only qualifying pair that does not include E and the one the 34% threshold was chosen to test.
• (c)6 — One too many — it would need A+U at 32% to qualify, but 32 falls short of 34. A+U and A+I differ by only three points, which is what makes the threshold do real work.
• (d)7 — Three too many. It would require A+O at 28%, I+O at 23% or I+U at 27% to clear 34%, and none of them comes close.

For mutually exclusive outcomes the probability of 'one or the other' is the sum of the individual probabilities: P(A or B) = P(A) + P(B). A single vowel occurrence drawn from the book is exactly one of the five, so the events do not overlap and no subtraction of an intersection is needed. The counting side is equally simple — the number of unordered pairs from five items is 5C2 = 10.

The efficient way to work this is not to compute all ten sums at all. Sort the percentages: E 45, A 20, I 15, U 12, O 8. Since E alone is 45, every pair with E is already above 34, which gives four. Now ask which pairs among the remaining four vowels can reach 34: the largest such pair is A+I = 20 + 15 = 35, which qualifies, and the next largest is A+U = 32, which does not. Everything below that is smaller still, so the count stops at five. That takes about fifteen seconds and is far less error-prone than a table of ten additions under exam pressure. Two small points of care: the question says 'more than 34%', so a sum of exactly 34 would not count, and the phrase 'any one of the two' means either vowel, which is the union and therefore a sum, not a product.

• The shared data set: A 20%, E 45%, I 15%, O 8%, U 12% — the five figures add to 100.
• Number of unordered pairs from five vowels = 5C2 = 10.
• For mutually exclusive events, P(one or the other) = sum of the two probabilities.
• The ten pair-sums: 65, 35, 28, 32, 60, 53, 57, 23, 27, 20.
• Five sums exceed 34 — the four pairs containing E, plus A+I at 35.

• Counting only the pairs containing the largest value and missing A+I.
• Treating 'more than 34%' as 'at least 34%' — here it makes no difference, but in a set where a sum lands exactly on the threshold it decides the answer.
• Multiplying the two percentages instead of adding them, which would answer a different question entirely.

Asked as the first item of a three-question data set, where the table is printed once and the arithmetic per item is small but the counting must be exhaustive.

• practice — not a real PYQ

  Using the same distribution (A 20, E 45, I 15, O 8, U 12), for how many pairs of vowels is the combined chance of occurrence less than 30%?

  • (a)2
  • (b)3
  • (c)4
  • (d)5

  Answer(c) 4 — A+O 28, I+O 23, I+U 27 and O+U 20 all fall below 30; A+U at 32 does not.

• practice — not a real PYQ

  If a vowel is picked at random from the same book, the chance that it is either A or O is

  • (a)12%
  • (b)20%
  • (c)28%
  • (d)160%

  Answer(c) 28% — the two events are mutually exclusive, so add: 20 + 8 = 28.