EPFO APFC 2023 General Studies · Question 74 of 120

Correct — D, (d) 1/5, printed in the booklet as 1 over 5. The expression factorises, and that is the whole problem: xy plus x plus y equals (x plus 1)(y plus 1) minus 1. Adding 1 to both sides, the expression is even exactly when (x plus 1)(y plus 1) is odd. A product is odd only when every factor is odd, so both x plus 1 and y plus 1 must be odd, which means both x and y must be even. The same conclusion follows without the factorisation by checking the three parity cases directly. If both are even, xy is even and x plus y is even, so the whole expression is even. If one is even and the other odd, xy is even but x plus y is odd, so the expression is odd. If both are odd, xy is odd and x plus y is even, so the expression is odd again. Only the both-even case works. Now count. The set has three even members, 2, 4 and 6, and three odd ones. Taking the two values as an unordered pair of distinct numbers, the total number of pairs is 6 choose 2, which is 15, and the number in which both are even is 3 choose 2, which is 3, giving 3 over 15, that is 1/5. Counting ordered pairs instead gives 6 times 5, or 30, in total and 3 times 2, or 6, favourable, which is 6 over 30 and the same 1/5. Ordered and unordered counting must agree here because the condition treats x and y symmetrically, and checking that they do is a cheap way of confirming that nothing has been double counted.

• (a)1/2 — One half is the reflex answer to any question about whether something is odd or even, on the loose feeling that half of all numbers are even. It is worth seeing that no quantity in this problem equals one half. The probability that x plus y is even is 6 over 15, that is two fifths, since both must be even or both odd. The probability that the product xy is even is 12 over 15, or four fifths, since the only way to avoid it is for both to be odd. And the probability the question actually asks about is 1/5. Parity questions are not coin flips once a constraint links the two variables, and this one links them tightly: the expression is even only in a single one of the three parity cases, and that case is the least likely of the three when distinct values are drawn.
• (b)1/3 — One third is what a candidate gets by identifying the three parity cases correctly — both even, one of each, both odd — establishing that exactly one of them makes the expression even, and then treating the three cases as equally likely. They are not. Among the 15 unordered pairs of distinct values, 3 have both even, 9 have one of each and 3 have both odd, so the mixed case is three times as common as either of the others. This is one of the most frequent errors in elementary probability: the number of cases is not the same as their probability, and only when the cases are equally likely can one be divided by the other. Whenever a solution ends by dividing one by the number of categories, check whether the categories are of equal size before accepting it.
• (c)1/4 — One quarter is the exact answer to the same question with the word distinct removed. If x and y were allowed to take the same value, each would independently be even with probability three sixths, and the two conditions being independent, the probability that both are even would be one half times one half, which is one quarter. The presence of this value in the option set is a signal that the distinctness condition is the point of the question and not decoration. Distinctness matters because it makes the two draws dependent: once x has taken an even value, only two of the five remaining values are even, so the second probability is two fifths rather than one half, and the product is three sixths times two fifths, which is 1/5. Read the sampling condition — with or without replacement, distinct or not — before writing any probability down.

Two ideas meet in this question. The first is parity algebra, the practice of reducing a statement about odd and even numbers to a statement about factors. The identity xy plus x plus y equals (x plus 1)(y plus 1) minus 1 is the standard manipulation for expressions of this shape, and it converts a question about the parity of a sum of three terms into a question about the parity of a single product, which is easy because a product is odd only when all its factors are odd. Where such an identity is not visible, the same result follows from a small parity table listing the three combinations of odd and even inputs, and building that table is never more than three lines of work. The second idea is counting under a distinctness constraint. Drawing two distinct values from a set of six is sampling without replacement, so the two draws are not independent: the first even value leaves only two evens among five remaining numbers. The count can be done with unordered pairs, using combinations, or with ordered pairs, using the product rule, and the two must agree whenever the event is symmetric in the two variables. Doing it both ways is the standard self-check, and it costs almost nothing on a set this small.

The quantitative block of this paper sets one probability item in most groups, and the numbers involved are always small enough for exhaustive counting to be feasible — six values here, thirty ordered pairs — so no candidate is ever forced to rely on a remembered formula. What separates the candidates is whether the condition is translated correctly before the counting begins. Here the translation is the entire difficulty: once the condition becomes both x and y even, the counting is a single line. The option set is built around the three ways of getting the translation or the sampling wrong, and each wrong option is the exact answer to a slightly different question — one half to a question with no constraint at all, one third to a question in which the three parity cases are taken as equally likely, and one quarter to the same question with repetition allowed. That is a common construction in this paper and a useful diagnostic when checking work: if your answer is one of the options, that does not confirm it, because the wrong options are the answers to the near-miss questions. Reread the stem and satisfy yourself that the question you solved is the one printed. Note finally that the booklet prints all four options as stacked fractions, reproduced here inline with a solidus.

• The expression xy plus x plus y equals (x plus 1)(y plus 1) minus 1, so it is even exactly when (x plus 1)(y plus 1) is odd, which requires both x and y to be even.
• A parity table confirms it: both even gives an even result, one even and one odd gives an odd result, and both odd gives an odd result, so only one of the three cases succeeds.
• Among the numbers 1 to 6 there are three even values, so the number of unordered pairs of distinct values with both even is 3 choose 2, that is 3, out of a total of 6 choose 2, that is 15, giving 1/5.
• Counting ordered pairs gives 6 times 5, or 30, in total and 3 times 2, or 6, favourable, which is again 1/5; ordered and unordered counting agree because the event is symmetric in x and y.
• If the values were allowed to repeat, the two draws would be independent and the probability that both are even would be one half times one half, that is one quarter — which is why the distinctness condition changes the answer.

• Treating the three parity cases as equally likely; among distinct pairs there are 3 both-even, 9 mixed and 3 both-odd, so the counts are 3, 9 and 3 and not equal thirds
• Overlooking the word distinct, which turns sampling without replacement into sampling with replacement and changes the answer from 1/5 to 1/4
• Assuming any parity question has probability one half; here the probability that the sum is even is two fifths and that the product is even is four fifths, and neither is a half
• Mixing ordered and unordered counting between numerator and denominator, which is the usual source of answers that are out by a factor of two
• Accepting an answer merely because it appears among the options; each wrong option here is the correct answer to a slightly different question

Probability in this paper is set on small, fully enumerable sample spaces — a set of six numbers, two dice, a handful of letters or coins — and the item is decided by the translation of a verbal condition into a countable event rather than by any advanced technique. Expressions whose parity or divisibility must be examined are a favourite vehicle, because they let the setter hide a simple counting problem behind an algebraic surface. The reliable method is the same every time: simplify or factorise the expression until the condition becomes a statement about the inputs, list the cases, count them under the sampling rule the stem specifies, and then verify by counting a second way. Where the numbers are small enough, an outright enumeration of the sample space is a legitimate and often faster method, and it is worth practising so that it can be done quickly and without transcription errors.

No directly related past PYQ was found.

• practice — not a real PYQ

  Two distinct numbers x and y are chosen from the set {1, 2, 3, 4, 5, 6}. What is the probability that the product xy is odd?

  • (a)1/5
  • (b)1/4
  • (c)1/3
  • (d)1/2

  Answer(a) 1/5 — a product is odd only when both factors are odd, and the set has three odd members, so the favourable count is 3 choose 2, that is 3, out of 6 choose 2, that is 15. The answer coincides with the probability that both are even because the set contains equally many odd and even numbers; if the set were 1 to 7 the two probabilities would differ, which is worth checking as an exercise.

• practice — not a real PYQ

  For integers x and y, the expression xy + x + y is even if and only if

  • (a)both x and y are odd
  • (b)both x and y are even
  • (c)exactly one of x and y is even
  • (d)x and y have the same parity

  Answer(b) both x and y are even — since xy + x + y equals (x + 1)(y + 1) − 1, the expression is even exactly when (x + 1)(y + 1) is odd, and a product is odd only when both factors are odd, that is when both x and y are even. Note that having the same parity is not enough: if both are odd, the product is odd and the sum even, so the whole expression is odd.