Basic Probability

Overview

Basic Probability appears about 1 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Numerical Ability.

Basic Probability is one of the cleanest scoring topics in AFCAT Numerical Ability. The exam keeps probability at the favourable-over-total level. There is almost no calculus, no permutation-heavy combinatorics, and conditional probability shows up only as an afterthought. If you memorise four or five standard sample spaces — single die, two dice, one coin, two coins, three coins, and a 52-card pack — you can crack nearly every probability item you will see in under a minute.

Across recent AFCAT papers the topic appears almost every cycle. It contributes around 1 question per paper, which means three marks added or one mark lost. Because the wording is short and the arithmetic is light, this is exactly the kind of item you should attempt early in the section rather than leave for the end. The challenge is not difficulty — it is avoiding small enumeration mistakes under time pressure.

Why probability is high-accuracy in AFCAT

Probability rewards candidates who slow down for ten seconds to list the sample space, then speed through the arithmetic. Unlike data interpretation or time-speed-distance, there are no traps hidden in the question wording itself. The trap, if any, is in the candidate skipping straight to a guess without writing favourable outcomes on paper.

• The sample space size is fixed and small — 6, 36, 52, or 4 — so total outcomes are never in doubt.
• The arithmetic stops at simple fraction reduction. No decimal conversion is needed because options are in fraction form.
• Most questions are direct one-step problems. Two-step problems usually involve drawing two balls without replacement.
• The same five or six patterns repeat across papers — sum of two dice, face card from a pack, both heads from two coins, at-least-one head from three coins, and so on.

Basics — sample space, event, and probability

Three terms anchor every probability problem. Once you can name them in a given question, the formula does itself.

• Experiment: the action being performed — tossing a coin, rolling a die, drawing a card.
• Sample space (S): the complete set of possible outcomes of the experiment. The number of elements in S is n(S).
• Event (E): a subset of the sample space — the outcomes we care about. The number of elements in E is n(E).
• Probability of E: P(E) = n(E) / n(S) = favourable outcomes / total outcomes.

If a die is rolled, the sample space is {1, 2, 3, 4, 5, 6} so n(S) = 6. If the event is rolling a prime number, the favourable outcomes are {2, 3, 5} so n(E) = 3. Therefore P(prime) = 3/6 = 1/2. That is the entire mechanism of AFCAT probability — the rest is just learning which sample space to write down.

Properties and identities you must memorise

These five identities cover every AFCAT probability question. Write them on the rough sheet at the start of the section.

Property	Formula	When it applies
Range	0 ≤ P(E) ≤ 1	Always. P(E) = 0 means impossible, P(E) = 1 means certain.
Complement	P(not E) = 1 − P(E)	Use for "at least one" problems.
Union of two events	P(A or B) = P(A) + P(B) − P(A and B)	Whenever the question says "A or B".
Mutually exclusive	P(A and B) = 0, so P(A or B) = P(A) + P(B)	When A and B cannot happen together — like a card being red and being a spade.
Independent events	P(A and B) = P(A) × P(B)	When the outcome of A does not affect B — like two tosses of one coin.

The most common slip is forgetting to subtract P(A and B) in the union formula when the events overlap. For example, P(card is red or king) is not P(red) + P(king); the two red kings would get counted twice.

Standard sample spaces to memorise

If you know the size of the sample space cold, you halve your solving time. These six come up again and again.

Experiment	Size of sample space	Notes
Toss one coin	2	{H, T}
Toss two coins	4	{HH, HT, TH, TT}
Toss three coins	8	2 × 2 × 2; useful for at-least-one questions
Roll one die	6	{1, 2, 3, 4, 5, 6}
Roll two dice	36	6 × 6; ordered pairs (i, j)
Draw one card	52	Standard pack

For three coins, the eight outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The number of heads ranges from 0 to 3, and the distribution is 1, 3, 3, 1 — a small Pascal pattern worth remembering for "exactly two heads" type items.

Two-dice sum and product tables

When two dice are thrown the sample space has 36 ordered pairs. The sum of the two faces ranges from 2 to 12, but not uniformly. The frequencies form a triangular distribution that peaks at 7.

Sum	Favourable pairs	Count	Probability
2	(1,1)	1	1/36
3	(1,2), (2,1)	2	2/36 = 1/18
4	(1,3), (2,2), (3,1)	3	3/36 = 1/12
5	(1,4), (2,3), (3,2), (4,1)	4	4/36 = 1/9
6	(1,5), (2,4), (3,3), (4,2), (5,1)	5	5/36
7	(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)	6	6/36 = 1/6
8	(2,6), (3,5), (4,4), (5,3), (6,2)	5	5/36
9	(3,6), (4,5), (5,4), (6,3)	4	4/36 = 1/9
10	(4,6), (5,5), (6,4)	3	3/36 = 1/12
11	(5,6), (6,5)	2	2/36 = 1/18
12	(6,6)	1	1/36

Other common two-dice events:

• Doubles (both faces equal): 6 favourable — (1,1) to (6,6). P = 6/36 = 1/6.
• Sum greater than 9: sums 10, 11, 12 → 3 + 2 + 1 = 6. P = 6/36 = 1/6.
• Sum is a prime number: primes between 2 and 12 are 2, 3, 5, 7, 11. Counts 1 + 2 + 4 + 6 + 2 = 15. P = 15/36 = 5/12.
• Product is even: easier through the complement. Product is odd only when both faces are odd — 3 × 3 = 9. So P(odd) = 9/36 = 1/4 and P(even) = 3/4.

Composition of a 52-card pack

Card problems are almost always one-card draws in AFCAT. Memorise the breakdown so you never count cards in the exam hall.

Category	Count	Probability of single draw
Total cards	52	1
Suits	4 (spades, hearts, diamonds, clubs)	13/52 = 1/4 per suit
Red cards (hearts + diamonds)	26	26/52 = 1/2
Black cards (spades + clubs)	26	26/52 = 1/2
Face cards (J, Q, K of each suit)	12	12/52 = 3/13
Aces	4	4/52 = 1/13
Kings, queens, or jacks individually	4 each	4/52 = 1/13 each
Number cards (2 through 10)	36	36/52 = 9/13
Red kings	2	2/52 = 1/26

Ball drawing — favourable counting with constraints

Bag-of-balls questions are the second most common probability format in AFCAT. The bag holds balls of two or three colours; the question asks for the probability of a stated colour pattern when one or more balls are drawn.

The total outcomes use the combination C(n, r) — the number of ways to choose r balls from n. For AFCAT you only need small values:

• C(n, 1) = n
• C(n, 2) = n(n − 1) / 2
• C(n, 3) = n(n − 1)(n − 2) / 6

Pattern 1 — both same colour. Bag has a red and b blue. Two balls are drawn. P(both red) = C(a, 2) / C(a + b, 2). P(both blue) = C(b, 2) / C(a + b, 2). P(both same colour) = sum of the two.

Pattern 2 — both different colour. Favourable = a × b (one red and one blue, in any order, treated as a combination). P = ab / C(a + b, 2).

Pattern 3 — at least one of a colour. Use the complement. P(at least one red) = 1 − P(no red) = 1 − C(b, r) / C(a + b, r), where r is the number drawn.

With replacement versus without replacement

Read the question carefully — the words "without replacement" or "the first ball is not put back" change the denominator.

Scenario	What changes	Example denominators (bag of 10)
With replacement	Total stays the same on each draw. Events are independent.	1st draw: 10, 2nd draw: 10. Total ordered outcomes: 100.
Without replacement	Total drops by 1 on the second draw. Events are dependent.	1st draw: 10, 2nd draw: 9. Total ordered outcomes: 90.

If a bag has 4 red and 6 blue balls and two balls are drawn:

• With replacement, both red: (4/10) × (4/10) = 16/100 = 4/25.
• Without replacement, both red: (4/10) × (3/9) = 12/90 = 2/15.

The without-replacement answer is always smaller because the first red removes a red from the bag, shrinking the favourable count on the second draw.

Conditional probability — a light touch

Conditional probability rarely appears in AFCAT, but a single question may show up once every few cycles, usually framed as "given that X has happened, find the probability of Y".

Formula: P(A given B) = P(A and B) / P(B), provided P(B) is not zero.

In plain English, conditioning on B means we have shrunk the sample space to just those outcomes where B is true. The numerator counts outcomes where both A and B happen; the denominator counts outcomes where B happens.

Quick example: A die is rolled. Given that the outcome is even, what is the probability that it is greater than 3? The conditioning shrinks the sample space to {2, 4, 6}. Within that set, outcomes greater than 3 are {4, 6}. P = 2/3.

That is all the depth AFCAT needs. Do not invest more than five minutes of prep here.

Common AFCAT trap patterns

1. Forgetting the overlap subtraction in OR problems. P(red card or king) is not 26/52 + 4/52. The two red kings are double-counted, so subtract 2/52. Correct answer: 28/52 = 7/13.
2. Reading "at least one" as "exactly one". At least one means one or more — handle it through the complement: 1 − P(none).
3. Mixing replacement and no-replacement. Always re-read the question after solving. If the wording says "the ball is not replaced" and you used the same denominator for both draws, you have over-counted.
4. Ordered versus unordered pairs in two-dice problems. The 36-outcome sample space is ordered, so (3, 4) and (4, 3) are distinct. Treat the dice as distinguishable even if they look identical.
5. Counting the joker. Standard AFCAT problems use a 52-card pack with no jokers. If the question explicitly says "pack of 54 cards including jokers", adjust both numerator and denominator.
6. Confusing "at most" with "at least". At most 2 means 0, 1, or 2; at least 2 means 2 or more. The complements differ.

Time budget for the section

AFCAT Numerical Ability has 18 to 22 questions in roughly 33 to 36 minutes. The benchmark is about 100 seconds per item, but you should bank time on easy topics like probability so that you can spend extra time on data interpretation or mixtures.

Probability sub-pattern	Target time
Single die or single coin item	20–30 seconds
Single card draw	30 seconds
Two-dice sum item	40–50 seconds
Ball drawing without replacement	60–75 seconds
OR with overlap, or at-least-one	60–90 seconds

If you have memorised the two-dice sum table and the card-pack composition, you should comfortably finish any AFCAT probability item inside 60 seconds.

Worked AFCAT-style examples

Exam-day strategy

1. Write down n(S) the moment you read a probability question. If the total outcomes are unclear, you cannot trust the answer.
2. For any 'at least one' problem, default to the complement: 1 − P(none). It is faster than summing the cases of exactly one, exactly two, and so on.
3. When two events are joined by 'or', check whether they can both happen together. If yes, subtract the intersection.
4. If the question involves two draws, look for the words 'with replacement' or 'without replacement' before computing.
5. Reduce the final fraction. AFCAT options are always given in lowest terms, so 4/8 will not appear — 1/2 will.
6. Memorise the sample-space sizes (2, 4, 6, 8, 36, 52) and the two-dice sum frequencies (1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1).
7. Aim for 45 to 60 seconds per probability item. If you stall, mark and revisit.

Frequently asked questions

How many probability questions appear in AFCAT?

On average about 1 question per paper from basic probability, sometimes two when the Numerical Ability section is light on data interpretation.

Is conditional probability tested in AFCAT?

Very rarely. When it does appear it is restricted to one-line questions solvable by shrinking the sample space, not Bayes' theorem.

Do I need to learn permutations and combinations for AFCAT probability?

Only the basics — C(n, 1), C(n, 2), and occasionally C(n, 3) for ball-drawing problems. There are no large factorial computations.

Are card and dice questions equally likely?

Cards and dice together account for the majority of AFCAT probability items. Ball-drawing problems come next, and coin-tossing is the least frequent.

What is the most common mistake candidates make?

Forgetting to subtract the intersection in 'A or B' questions when the two events overlap — for example, treating king and heart as if they were mutually exclusive.

Should I memorise the two-dice sum table?

Yes. The frequencies 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2 through 12 save you 30 seconds per two-dice item.

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