There are four identically sized and shaped balls in a box, with only its top open. Each ball is of a different colour, these being : Green, Red, White and Blue, only one ball of each colour. Without looking into the box, one ball is randomly picked out and its colour is noted; then it is returned to the box. What are the chances that, in two successive draws, one may get the white ball and the red ball — in whichever order ?
- (a)1/16
- (b)1/8
- (c)1/4
- (d)1/2
Answer
Why
Correct — B, (b) 1/8.
Two clauses in the stem do all the work, and both are easy to read past. The ball is RETURNED TO THE BOX after the first draw, and the two colours may come up IN WHICHEVER ORDER.
BECAUSE THE BALL IS RETURNED, the two draws are independent and the box is identical on both occasions: four balls, one of each colour, every colour a probability of 1/4 every time. Nothing depletes.
So list the ordered outcomes. Draw one has 4 possibilities, draw two has 4, giving 4 × 4 = 16 equally likely ordered pairs. How many of them consist of the white ball and the red ball ?
(white, red) and (red, white)
Two of the sixteen. The probability is 2/16 = 1/8, which is option (b).
The same figure by multiplication:
P(white then red) = 1/4 × 1/4 = 1/16 P(red then white) = 1/4 × 1/4 = 1/16 the two are mutually exclusive, so add: 1/16 + 1/16 = 1/8
THE ORDER CLAUSE IS THE WHOLE ITEM. 'In whichever order' means two distinct sequences satisfy the requirement, so the single-sequence probability has to be doubled. An unordered requirement always covers more outcomes than an ordered one, and the factor is exactly the number of arrangements — here 2! = 2 for two distinct colours.
WHAT REPLACEMENT CHANGES, worth seeing once. Had the ball not been returned, the second draw would be from three balls and the answer would be 2 × (1/4 × 1/3) = 1/6, which is not on offer at all. The clause is load-bearing rather than decorative.
The four fractions are printed inline with a solidus rather than stacked, and the stem uses an em dash before its final phrase; the colour names are capitalised in the list and lower-case later in the same sentence, all as the booklet sets them.
Why the others are wrong
- (a)1/16 — The probability of ONE specified order — white first and then red — and therefore the answer to a question the paper did not ask. It is the single most likely error on the item, because 1/4 × 1/4 = 1/16 is the calculation a candidate reaches for first and it looks finished. What it leaves out is the phrase 'in whichever order', which admits the reverse sequence too. Two of the sixteen ordered pairs qualify, not one. The general form of the mistake is treating an unordered requirement as an ordered one, and the correction is always the same: multiply by the number of arrangements of the required outcomes, which for two distinct items is 2.
- (c)1/4 — The probability of drawing one specified colour on ONE draw, which is where a candidate lands who computes the chance of the white ball and then stops without folding in the second draw at all. It is also the value that tempts anyone who reasons loosely that 'two of the four colours are wanted, so it must be about a quarter' — but that argument confuses which two things are being counted. Note that 1/4 is four times the correct answer, and the question asks about a compound event across two draws; a compound requirement is always LESS likely than either of its parts, so any answer as large as the single-draw probability should be suspect before it is computed.
- (d)1/2 — This is the probability that a SINGLE draw produces one or other of the two named colours — white or red, two of the four balls, so 2/4 = 1/2. It answers a much weaker question than the one asked, since getting 'one of the two' on one draw says nothing about getting both across two draws. It is the largest value offered and the easiest to rule out on general grounds: requiring two specified outcomes in two draws must be considerably less likely than requiring one acceptable outcome in one draw. Checking the plausibility of the magnitude before doing any arithmetic disposes of this option and of option (c) together.
Concept
This is a problem in COMPOUND PROBABILITY, and it is decided by three questions asked in order: is there replacement, does order matter, and are the favourable outcomes mutually exclusive ?
WITH OR WITHOUT REPLACEMENT. When the item drawn is returned, the population is restored, successive draws are INDEPENDENT, and each probability stays the same on every draw. When it is not returned, the draws are dependent and each subsequent probability must be conditioned on what has already gone — the denominator shrinks. Here the ball is returned, so 1/4 applies twice.
DOES ORDER MATTER. A requirement stated as a sequence ('white first, then red') is satisfied by one arrangement; a requirement stated as a set ('one white and one red, in whichever order') is satisfied by all arrangements of that set. For two distinct items there are 2! = 2 arrangements, so the set version is twice as likely as the sequence version. This is the same combinatorial idea that separates permutations from combinations, met here in probability form.
MULTIPLY OR ADD. Multiply probabilities to combine events that must ALL happen — white on the first draw AND red on the second. Add probabilities to combine mutually exclusive alternatives, EITHER of which would do — (white, red) OR (red, white). Almost every compound-probability item is one application of each, in that order: multiply within a sequence, add across sequences.
THE COUNTING ROUTE IS THE SAFEST. Because every ordered pair here is equally likely, the probability is simply favourable outcomes over total outcomes: 2 over 16. Setting a problem up as a count of equally likely outcomes removes any need to decide when to multiply and when to add, and it makes the sample space visible — a decisive advantage when the wording is ambiguous.
A STANDARD CHECK ON MAGNITUDE. The probability of a compound requirement is never greater than the probability of any single component. So an answer for 'both colours across two draws' cannot exceed the 1/4 chance of one colour on one draw. On this item that check alone eliminates half the option set.
Quantitative aptitude is the largest strand on this APFC paper, and probability appears within it as a short worded item rather than as a formula exercise. The arithmetic is trivial by design; what is being tested is careful reading.
This stem is unusually generous about the fact that reading is the test. It states explicitly that the balls are identically sized and shaped, that there is exactly one ball of each colour, that the drawer cannot see into the box, and that the ball is returned — four clauses whose only purpose is to establish that all four outcomes are equally likely and that the draws are independent. When a probability stem spends that much space on the physical set-up, it is telling you which assumptions the calculation rests on.
Then it appends the clause that the option set is built around, 'in whichever order', set off by an em dash at the very end of a long sentence, where it is easiest to skim past. The value produced by ignoring it is on the page as an option. That is the recurring architecture of probability items on these papers: the wrong answers are not miscalculations but answers to slightly different questions — one order instead of two, one draw instead of two, either colour instead of both.
The habit worth building is to restate the requirement in your own words before computing, naming whether replacement occurs and whether order matters. On an item with a four-element sample space it is even quicker to write the sixteen ordered pairs out mentally and count. Enumeration scales badly, but where the sample space is small it is exact, and it makes the ambiguity in the wording impossible to overlook.
Key facts
- With replacement the draws are independent and every colour keeps a probability of 1/4 on every draw; without replacement the second draw would be from three balls.
- Two draws from four colours give 4 × 4 = 16 equally likely ORDERED outcomes.
- 'In whichever order' admits both (white, red) and (red, white), so two of the sixteen outcomes qualify and the probability is 2/16 = 1/8.
- Multiply probabilities for events that must all occur within one sequence; add probabilities across mutually exclusive sequences.
- An unordered requirement is more likely than the corresponding ordered one by exactly the number of arrangements — a factor of 2! = 2 for two distinct items.
- Had the ball not been replaced the answer would be 2 × (1/4 × 1/3) = 1/6, a value the paper does not offer, so the replacement clause is doing real work.
- A compound event can never be more probable than any of its component events — a magnitude check that rules out two of the four options here at sight.
Study next
Common traps
- Missing the words 'in whichever order' and answering with the single-sequence probability, which the paper prints as an option.
- Missing the replacement clause and reducing the denominator on the second draw; here that would give 1/6, which is not even on offer.
- Adding where multiplication is needed, or the reverse. Multiply within a sequence of events that must all occur; add across alternative sequences.
- Answering with the probability for one draw when the question spans two, which is how the value 1/4 gets chosen.
- Confusing 'one of the two named colours' with 'both of the two named colours' — the first gives 1/2 on a single draw and answers a much weaker requirement.
Probability on EPFO papers is set as a short story problem with a small, fully specified sample space — coloured balls in a box, a coin tossed twice, dice thrown twice, cards drawn from a pack — and the computation is deliberately light. What varies between items is a single clause: with or without replacement, in a stated order or in any order, exactly one or at least one. The distractors are then generated by dropping that clause rather than by miscalculating, so every wrong option is the correct answer to a neighbouring question. The reliable defence is to restate the requirement before computing, decide explicitly whether the draws are independent and whether order matters, and where the sample space is small enough, enumerate it and count.
Related PYQs
EPFO_APFC_2023_Q74Suppose that x and y are distinct variables that take values from {1, 2, 3, 4, 5, 6}. What is the probability that the value of the expression xy + x + y is even?
- (a) 1/2
- (b) 1/3
- (c) 1/4
- (d) 1/5
Answer(d) 1/5
A probability item on the later APFC paper, again solved by counting equally likely ordered pairs rather than by reaching for a formula.
EPFO_EOAO_2023_Q40A dice is thrown two times. The number of ways that the number appearing on the first throw is not less than that on the second throw is :
- (a) 15
- (b) 20
- (c) 21
- (d) 36
Answer(c) 21
A dice-thrown-twice item that turns entirely on treating the outcomes as ORDERED pairs — the same distinction between a sequence and a set that decides this question.
EPFO_APFC_2016_Q90In a chess tournament, each of the six players will play with every other player exactly once. What is the number of matches that will be played during the tournament ?
- (a) 10
- (b) 15
- (c) 20
- (d) 25
Answer(b) 15
The chess-tournament item on this paper, where the counting runs the other way: pairings there are unordered, so the arrangements are divided out instead of multiplied in.
Practice
- practice — not a real PYQ
A fair coin is tossed twice. What is the probability of getting exactly one head ?
- (a)1/4
- (b)1/3
- (c)1/2
- (d)3/4
Answer(c) 1/2 — the four equally likely ordered outcomes are HH, HT, TH and TT, and two of them contain exactly one head, giving 2/4 = 1/2. The value 1/4 is the probability of one SPECIFIED order such as head then tail, which is the same order-blindness that the balls-in-a-box item is built around.
- practice — not a real PYQ
A box holds four balls, one green, one red, one white and one blue. Two balls are drawn one after the other and the first is NOT returned to the box. What is the probability of drawing the white ball and the red ball, in whichever order ?
- (a)1/12
- (b)1/8
- (c)1/6
- (d)1/4
Answer(c) 1/6 — without replacement the second draw is from three balls, so each order has probability 1/4 × 1/3 = 1/12, and the two admissible orders give 2/12 = 1/6. Equivalently, there are 4 × 3 = 12 ordered pairs and two of them qualify. The answer is larger than the 1/8 of the with-replacement case because removing a ball raises the chance that the second draw is the other wanted colour.