Probability Basics for Entrance Exams
Probability is the topic where confident students go wrong most often, because the arithmetic is easy and the traps are hidden in the wording. "Or" is not always addition. "And" is not always multiplication. "At least one" is almost never worth counting directly. This guide sets out the rules, works eight questions with every line shown, and checks two of them by a second method so you can see the two routes agree.
The definition everything rests on
0 ≤ P(A) ≤ 1 and P(not A) = 1 − P(A)
The words equally likely carry all the weight. Counting outcomes only works when each one is as probable as every other — which is why "two dice" questions are counted over the 36 ordered pairs, not over the 11 possible totals. The totals are not equally likely; the pairs are.
The four rules you will use
Mutually exclusive events: P(A and B) = 0, so P(A or B) = P(A) + P(B)
Multiplication: P(A and B) = P(A) × P(B | A)
Independent events: P(B | A) = P(B), so P(A and B) = P(A) × P(B)
Conditional: P(B | A) = P(A and B) ÷ P(A)
Counting problems often need combinations, since order usually does not matter when objects are drawn together:
Worked example 1 — a sum on two dice
Two fair dice are thrown. Find P(sum = 8).
Total equally likely outcomes = 6 × 6 = 36.
Favourable ordered pairs: (2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 pairs.
P = 5/36 ≈ 0.139
Note that (2,6) and (6,2) are two different outcomes but (4,4) is only one. Missing that asymmetry is the classic error here.
Worked example 2 — the addition rule with overlap
One card is drawn from a standard 52-card pack. Find P(king or heart).
P(king) = 4/52, P(heart) = 13/52, P(king and heart) = 1/52 — the king of hearts is counted in both lists, so it must be subtracted once.
P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13 ≈ 0.308
Simply adding 4/52 and 13/52 gives 17/52, which double-counts one card. The overlap term is not optional.
Worked example 3 — drawing without replacement
Two cards are drawn together from a pack. Find P(both are aces).
Sequential method: P(first ace) = 4/52. Once an ace is gone, 3 aces remain
among 51 cards, so P(second ace | first ace) = 3/51.
P = (4/52) × (3/51) = 12/2652 = 1/221 ≈ 0.00452
Cross-check by combinations:
Ways to choose 2 aces from 4 = 4C2 = (4 × 3) ÷ 2 = 6
Ways to choose any 2 cards from 52 = 52C2 = (52 × 51) ÷ 2 = 1326
P = 6 ÷ 1326 = 1/221 — the same answer by a completely different route.
Worked example 4 — coloured balls, both ways
A bag holds 5 red and 3 blue balls. Two are drawn without replacement. Find P(both red).
Combinations: 5C2 ÷ 8C2 =
10 ÷ 28 = 5/14 ≈ 0.357
Sequential: (5/8) × (4/7) = 20/56 = 5/14 ✔
Worked example 5 — "at least one", the complement trick
A fair coin is tossed three times. Find P(at least one head).
Counting directly means adding the cases of exactly one, exactly two and exactly three heads. Much faster: the opposite of "at least one head" is "no heads at all".
P(no heads) = (1/2)³ = 1/8
P(at least one head) = 1 − 1/8 = 7/8 = 0.875
Whenever a question says "at least one", reach for 1 − P(none) first.
Worked example 6 — the binomial distribution
A fair coin is tossed five times. Find P(exactly 3 heads).
Here n = 5, r = 3, p = 1/2, q = 1/2.
5C3 = (5 × 4 × 3) ÷ (3 × 2 × 1) = 10
p³ q² = (1/2)³ × (1/2)² = (1/2)⁵ = 1/32
P = 10 × (1/32) = 10/32 = 5/16 = 0.3125
Worked example 7 — conditional probability
A die is thrown and you are told the result is even. Find the probability it is a 6.
The information shrinks the sample space from {1,2,3,4,5,6} to {2,4,6}.
P(6 | even) = P(6 and even) ÷ P(even) = (1/6) ÷ (3/6) = 1/3
Note it is not 1/6. Conditional probability means the denominator changes, and forgetting to change it is the standard mistake.
Worked example 8 — two bags, and working backwards
Bag A holds 3 red and 2 white balls. Bag B holds 1 red and 4 white balls. A bag is chosen at random and one ball is drawn from it. (i) Find P(red). (ii) Given the ball is red, find the probability it came from bag A.
(i) P(red) = P(A) × P(red | A) + P(B) × P(red | B)
= (1/2)(3/5) + (1/2)(1/5) = 3/10 + 1/10 = 4/10 = 2/5 = 0.4
(ii) P(A | red) = P(A and red) ÷ P(red) = (3/10) ÷ (4/10) = 3/4 = 0.75
The second part reverses the conditioning — this is Bayes' theorem in its simplest form, and it is a favourite in entrance papers precisely because the two questions look alike but have different answers.
Reading the question: which rule to use
| Wording in the question | What it usually means | Rule to reach for |
|---|---|---|
| "A or B" | Either event, possibly both | P(A) + P(B) − P(A and B) |
| "A and B", drawn together or in sequence | Both must happen | P(A) × P(B | A) |
| "With replacement" | The pool is restored, so trials are independent | P(A) × P(B) |
| "Without replacement" | The pool shrinks — not independent | P(A) × P(B | A) |
| "At least one" | Everything except the empty case | 1 − P(none) |
| "Given that…" | The sample space has already shrunk | P(A and B) ÷ P(given event) |
| "Exactly r out of n" | Repeated independent trials | Binomial formula |
Common mistakes that cost marks
- Adding probabilities of events that overlap. P(A or B) = P(A) + P(B) only when the two cannot happen together. Check for the overlap before you add.
- Treating "without replacement" as independent. The second denominator must drop: 4/52 then 3/51, never 4/52 twice.
- Counting unequal outcomes as equally likely. With two dice there are 36 equally likely pairs, not 11 equally likely totals.
- Forgetting to shrink the sample space in conditional questions. "Given the number is even" changes the denominator from 6 to 3.
- Using permutations where order does not matter. Cards drawn together are a combination problem; positions in a queue are a permutation problem.
- Not sanity-checking the final number. Any answer above 1 or below 0 is definitely wrong, and mutually exclusive probabilities over a complete set must sum to exactly 1.
- Grinding through "at least one" case by case. It is slower and it is where a case gets missed under time pressure.
Where probability appears in exams
| Exam | Typical use |
|---|---|
| CBSE / ICSE Class 11–12 | Addition and multiplication theorems, conditional probability, Bayes' theorem, binomial distribution |
| JEE | Multi-step counting combined with combinations and permutations |
| Aptitude sections of competitive papers | Quick single-step questions on dice, cards and coloured balls |
| Chemistry and physics later on | The Boltzmann distribution, radioactive decay and quantum probability all rest on these ideas |
For exact question patterns and the marks allotted, always check the current official notification and syllabus for your exam rather than relying on what a coaching handout said last year — these details do change between sessions.
Get the counting part right instantly. Most probability slips are actually nCr slips. The nPr / nCr calculator evaluates both permutations and combinations, so you can confirm a value like 52C2 = 1326 before building a whole answer on it.
Open the nPr / nCr Calculator →Preparing for boards and an entrance exam together? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — details at abcchemistry.in. If you are in Delhi, Noida or Gurgaon and want one-to-one help at home with the maths behind these questions, delhihometutor.com arranges home tuition across Delhi NCR.