Probability tells us how likely an event is, on a scale from 0 (impossible) to 1 (certain). In the NDA written exam it is a guaranteed scoring area — expect 4 to 6 questions every year on dice, cards, coins, conditional probability and Bayes’ theorem. The good news? The same handful of rules solve almost everything.
Why Probability Is a Must-Score Topic
Probability appears in Paper I (Mathematics) of the NDA exam, which carries 300 marks for 120 questions. Year after year, the probability and probability-distribution block delivers 4 to 6 questions — that is roughly 12 to 18 marks within easy reach.
What makes it special is that the questions are formula-driven. Once you know the rule, the arithmetic is light. There is no heavy calculation like in calculus, and no diagram-heavy reasoning like in geometry.
Probability answers are always between 0 and 1. If your final answer is negative or greater than 1, you have made a mistake — recheck immediately.
In this lesson we build from the absolute basics to Bayes’ theorem and the binomial distribution, with worked examples in the exact style the NDA loves.
Probability also rewards speed. Because each question is short, a candidate who has the rules memorised can clear the whole block in under ten minutes, leaving extra time for the longer trigonometry and calculus problems. Treat this chapter as your time-bank for the rest of the paper.
Sample Space, Outcomes and Events
Before any formula, learn the vocabulary — the NDA tests these definitions directly.
- Random experiment: an action whose result cannot be predicted with certainty, e.g. tossing a coin.
- Outcome: a single possible result of the experiment.
- Sample space (S): the set of all possible outcomes. For one die, S = {1, 2, 3, 4, 5, 6}.
- Event: any subset of the sample space, e.g. ‘getting an even number’ = {2, 4, 6}.
Types of events you must know
- Sure event: equals the whole sample space; probability = 1.
- Impossible event: the empty set ∅; probability = 0.
- Mutually exclusive events: cannot happen together, so A ∩ B = ∅.
- Exhaustive events: together they cover the entire sample space.
- Equally likely events: each outcome has the same chance.
- Complementary events: event E and its complement E′ together form the whole sample space and have no overlap.
Getting comfortable with this language pays off instantly, because most NDA questions are really just asking you to identify the event type and then apply the matching formula. For instance, the moment you spot the words ‘mutually exclusive’ you know the overlap term vanishes and the addition theorem becomes a simple sum.
Two coins give S = {HH, HT, TH, TT} — 4 outcomes, not 3. Order matters, so HT and TH are different. Forgetting this is the single most common slip in coin problems.
The Classical Probability Formula
When all outcomes are equally likely, probability is just counting.
P(E) = (Number of favourable outcomes) ÷ (Total number of outcomes in S)
That is, P(E) = n(E) ÷ n(S), where 0 ≤ P(E) ≤ 1.
The probability that event E does not happen is its complement:
P(E′) = 1 − P(E)
The complement rule is a huge time-saver. When a question says ‘at least one’, it is almost always faster to find P(none) and subtract from 1.
Notice that the classical formula only works when outcomes are equally likely. A loaded die or a biased coin breaks this assumption, and then you must be given the individual probabilities directly. The NDA usually keeps things fair, but always read the question to confirm before you start counting.
For a standard deck of 52 cards: 26 red, 26 black, 4 suits of 13 each, 12 face cards (J, Q, K) and 4 aces. Memorise these counts — card questions are very frequent.
The Addition Theorem (OR Events)
Use the addition theorem when a question asks for the probability of A or B happening.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
If A and B are mutually exclusive, then P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).
We subtract P(A ∩ B) so the overlap is not counted twice. This is exactly the inclusion–exclusion idea from set theory, and it extends to three events as well, where you add the singles, subtract the pairs, and add back the triple intersection.
Quick illustration
Draw one card from a deck. Find P(king or heart). There are 4 kings and 13 hearts, but the king of hearts is counted in both.
Students add 4/52 + 13/52 = 17/52 and stop. You must subtract the overlap (king of hearts), giving (4 + 13 − 1) ÷ 52 = 16/52 = 4/13.
The Multiplication Theorem (AND Events)
Use multiplication when both A and B must happen.
P(A ∩ B) = P(A) × P(B | A)
If A and B are independent, then P(B | A) = P(B), so P(A ∩ B) = P(A) × P(B).
Independent events: the outcome of one does not affect the other — like two separate coin tosses. Dependent events: the second probability changes because of the first — like drawing two cards without replacement.
Do not confuse mutually exclusive with independent. Mutually exclusive events cannot both occur, so they are actually dependent. Two events with non-zero probabilities cannot be both mutually exclusive and independent.
A bag has 5 red and 3 black balls. Two balls are drawn one after another without replacement. Find the probability that both are red.
Conditional Probability
Conditional probability is the chance of A given that B has already happened. It shrinks the sample space to B.
P(A | B) = P(A ∩ B) ÷ P(B), provided P(B) ≠ 0
Think of it as ‘zooming in’ on the region B and asking what fraction of it is also A.
A die is rolled. Given that the number is even, find the probability that it is greater than 3.
The word ‘given’ is your signal to use conditional probability. Reduce the denominator to the size of the ‘given’ event, then count favourables within it.
Total Probability and Bayes' Theorem
These two results handle problems with several boxes, bags or machines — a classic NDA favourite.
Theorem of Total Probability
If B1, B2, …, Bn are mutually exclusive and exhaustive events, then for any event A:
P(A) = P(B1)P(A | B1) + P(B2)P(A | B2) + … + P(Bn)P(A | Bn)
Bayes’ Theorem
Bayes’ theorem works backwards: given that A happened, which cause Bi was responsible?
P(Bi | A) = [ P(Bi)P(A | Bi) ] ÷ P(A)
where P(A) is found from the total-probability formula above.
P(Bi) are the prior probabilities (before the experiment) and P(Bi | A) are the posterior probabilities (after seeing the result A).
Random Variables and Expected Value
A random variable X assigns a number to each outcome — for example, the number of heads in 3 tosses. Its probability distribution lists each value of X with its probability, and these probabilities must add up to 1.
Mean / Expected value: E(X) = Σ xi P(xi)
Variance: Var(X) = E(X2) − [E(X)]2
A fair die is thrown once. X is the number shown. Find E(X).
So on average a fair die shows 3.5 — a value the die can never actually display, yet still the correct long-run mean. Expected value is best understood as the average result if the experiment were repeated thousands of times. The NDA often gives a small payout table (win rupees on some outcomes, lose on others) and asks for the expected gain; you simply multiply each amount by its probability and add, treating losses as negative.
The Binomial Distribution
When an experiment is repeated n independent times, each trial having only two outcomes — success (probability p) or failure (probability q = 1 − p) — the number of successes follows the binomial distribution.
P(X = r) = nCr pr qn−r, r = 0, 1, 2, …, n
Mean = np · Variance = npq · Standard deviation = √(npq)
For a binomial distribution, variance < mean always (because npq < np when q < 1). The NDA often tests this single fact directly.
A fair coin is tossed 4 times. Find the probability of getting exactly 2 heads.
Previous-Year Style Question
This problem combines total probability and Bayes’ theorem — a pattern that has appeared repeatedly in NDA papers.
Q. Bag I contains 3 red and 4 black balls; Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. What is the probability that it was drawn from Bag II?
Answer: Let B1, B2 be the events of choosing Bag I and Bag II, each with probability 1/2. P(red | B1) = 3/7 and P(red | B2) = 5/11. By Bayes’ theorem, P(B2 | red) = [(1/2)(5/11)] ÷ [(1/2)(3/7) + (1/2)(5/11)] = (5/11) ÷ (3/7 + 5/11) = (5/11) ÷ (68/77) = (35/77) ÷ (68/77) = 35/68.
In Bayes problems, the (1/2) priors cancel when the bags are equally likely. Spotting this lets you skip straight to comparing the conditional probabilities.
Common Traps and How to Dodge Them
Most lost marks in probability come from a small set of avoidable errors.
- With vs without replacement: always check whether the item is put back. It changes both the count and independence.
- ‘At least one’ problems: use P(at least one) = 1 − P(none) instead of adding many cases.
- Order in coins: n tosses give 2n equally likely outcomes — HT and TH are distinct.
- Confusing AND with OR: AND → multiply; OR → add.
For two dice, the sample space has 6 × 6 = 36 outcomes, not 12. Many students use 12 and get every two-dice question wrong. The smallest sum is 2 and the largest is 12.
Quick Revision
- Basic: P(E) = n(E)/n(S), with 0 ≤ P(E) ≤ 1 and P(E′) = 1 − P(E).
- Addition (OR): P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
- Multiplication (AND): P(A ∩ B) = P(A) × P(B | A); independent → P(A) × P(B).
- Conditional: P(A | B) = P(A ∩ B) / P(B).
- Bayes: P(Bi | A) = P(Bi)P(A | Bi) / P(A).
- Binomial: P(X = r) = nCr pr qn−r; mean = np, variance = npq.
Drill 15 to 20 mixed dice, card and bag problems and these rules will become automatic in the exam hall. A focused practice plan from The Cavalier mentors recommends solving one full set of previous-year probability questions every alternate day in the final month, timing yourself to under one minute per question. That discipline turns probability from a topic you hope to attempt into one you finish confidently with marks in the bank.
Frequently asked questions
How many probability questions come in the NDA Maths paper?
Typically 4 to 6 questions appear from probability and probability distributions in each NDA Mathematics paper, making it a reliable 12 to 18 mark scoring area.
What is the difference between mutually exclusive and independent events?
Mutually exclusive events cannot occur together (A ∩ B = empty), while independent events do not affect each other's probability. Two events with non-zero probability cannot be both at once.
When should I use Bayes' theorem?
Use Bayes' theorem when an event has already occurred and you must find the probability of which underlying cause (bag, machine or box) produced it — that is, a 'reverse' conditional probability.
What are the mean and variance of a binomial distribution?
For a binomial distribution with n trials and success probability p, the mean is np and the variance is npq, where q = 1 − p. Note that variance is always less than the mean.
How do I solve 'at least one' probability questions quickly?
Use the complement: P(at least one) = 1 − P(none). Finding the probability of none and subtracting from 1 is far faster than adding up many separate cases.
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