Probability

Probability is the backbone of quantitative reasoning in NDA Mathematics. Every NDA paper from 2010 through 2016 carries at least 6–8 questions on this topic, spanning classical counting problems, event relationships, conditional probability, and binomial distribution. The questions reward two things: clarity about definitions (mutually exclusive vs. independent events) and clean arithmetic on fractions. Master those, and you score full marks here.

This chapter also connects directly to Permutations and Combinations — most classical probability problems are solved by counting favourable outcomes using combinatorics. Build that chapter first if you are just starting.

What This Topic Covers

The NDA syllabus covers probability in two broad parts: classical probability theory and probability distributions. Within classical theory, you need to handle sample spaces, events, the addition theorem (for union of events), the multiplication theorem (for intersection), conditional probability, independence, and Bayes' theorem. Within distributions, the binomial distribution is the only one tested — its mean, variance, and the formula \(P(X = r) = \binom{n}{r} p^{r} q^{n-r}\).

Exam Pattern & Weightage

The table below is compiled from PYQ data across NDA papers from 2010 to 2016. Each row represents one examination paper (NDA I or NDA II for that year).

Year / Paper	No.	Subtopics Tested
2010-I	2	Independent events, calendar probability
2010-II	3	Classical probability, ball-drawing, independent events
2011-I	7	Coin+die, conditional, mutually exclusive, arrangement, Venn
2011-II	3	Circle geometry probability, letter arrangement
2012-I	5	Dice, calendar, binomial mode, two-dice events
2012-II	6	Certain event, urn problems, coin tosses, complement
2013-I	8	Binomial distribution, independent events, sample space, exhaustive
2013-II	7	Calendar, relay race, binomial parameters, conditional
2014-I	6	Ball drawing, alternating wins, Z = X+Y set problems
2014-II	6	Conditional, mutually exclusive/exhaustive, Bayes', arrangement
2015-I	7	Independent events, binomial mean/variance, hit-target
2015-II	8	Coin tosses, geometric probability, card drawing, binomial
2016-I	7	Dice sum, missile-hit binomial, conditional, calendar
2016-II	6	Independent events, cricket series, Bayes', three-part machine

Core Concepts

Classical Probability

Classical (Laplace) probability applies when all outcomes are equally likely. The sample space S is the set of all possible outcomes; an event E is a subset of S.

For three coins tossed simultaneously, \(n(S) = 2^{3} = 8\). The event "exactly 2 heads" has outcomes \(\{HHT, HTH, THH\}\), so \(n(E) = 3\) and \(P(E) = \tfrac{3}{8}\). This was asked in NDA 2012-II.

For two dice, \(n(S) = 36\). Sum = 7 occurs with outcomes \(\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\), so \(P(\text{sum} = 7) = \tfrac{6}{36} = \tfrac{1}{6}\). This was asked in NDA 2013-I.

Addition Theorem

For any two events \(A\) and \(B\):

Special case — mutually exclusive events: if \(A \cap B = \varphi\), then \(P(A \cap B) = 0\) and \(P(A \cup B) = P(A) + P(B)\). Two events \(A\) and \(B\) are exhaustive if \(A \cup B = S\), meaning \(P(A \cup B) = 1\). NDA 2013-II tested: "If \(A\) and \(B\) are exhaustive events, then their union is the sample space" — this is correct; but "their intersection must be empty" is not necessarily correct.

Multiplication Theorem

For independent events: \(P(A \cap B) = P(A) \cdot P(B)\). If \(A\) and \(B\) are independent, then so are \(A\) and \(B'\), \(A'\) and \(B\), and \(A'\) and \(B'\) — NDA 2010-II tested all three of these statements and all three are correct.

Conditional Probability

NDA 2011-I gave: \(P(A \cup B) = 0.5\), \(P(B) = 0.8\), \(P(A \mid B) = 0.4\). Find \(P(A \cap B)\). Since \(P(A \mid B) = \tfrac{P(A \cap B)}{P(B)}\), we get \(P(A \cap B) = 0.4 \times 0.8 = 0.32\). Then check: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\), so \(P(A) = 0.5 - 0.8 + 0.32 = 0.02\).

Bayes' Theorem

Bayes' theorem lets you update a probability given new information. It is used when you draw from one of several groups and need to find the probability that a specific group was selected given the outcome.

NDA 2014-II gave a classic Bayes' setup: there are 4 white and 3 black balls in Box 1, and 3 white and 4 black balls in Box 2. Roll a die — if \(\le 3\), draw from Box 2; otherwise from Box 1. Given the drawn ball is black, find the probability it came from Box 1. Apply Bayes' theorem with \(P(\text{Box1}) = P(\text{Box2}) = \tfrac{1}{2}\), \(P(\text{Black} \mid \text{Box1}) = \tfrac{3}{7}\), \(P(\text{Black} \mid \text{Box2}) = \tfrac{4}{7}\).

NDA 2015-II had a similar structure with an insurance company: 2000 scooter drivers, 4000 car drivers, 6000 truck drivers with accident probabilities 0.01, 0.03, and 0.15. One person meets an accident — find the probability the person is a scooter driver.

Binomial Distribution

The binomial distribution \(B(n, p)\) models the number of successes in \(n\) independent Bernoulli trials, each with success probability \(p\). Let \(q = 1 - p\).

NDA 2013-I asked: binomial distribution with \(n = 100\) and \(p = \tfrac{1}{3}\), find \(r\) where \(P(X = r)\) is maximum. \((n+1)p = \tfrac{101}{3} \approx 33.67\), so mode = 33. NDA 2013-I also established that the binomial distribution has exactly two parameters (\(n\) and \(p\)).

NDA 2015-I gave: mean = 2, variance = 1 for a binomial variate \(X\). Since mean \(= np = 2\) and variance \(= npq = 1\), we get \(q = \tfrac{1}{2}\), so \(p = \tfrac{1}{2}\) and \(n = 4\). Then \(P(X = 0) = \left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}\).

Worked Examples

Example 1 — Calendar Probability (2010-I, 2012-I)

Q: What is the probability that a leap year selected at random contains 53 Mondays?

• A leap year has 366 days = 52 complete weeks + 2 extra days.
• The 2 extra days can be any consecutive pair from the 7 possible pairs: \(\{\text{Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun}\}\).
• Pairs that include Monday: \(\{\text{Sun-Mon}\}\) and \(\{\text{Mon-Tue}\}\) — that is 2 favourable outcomes.
• $$P(\text{53 Mondays}) = \frac{2}{7}.$$ Answer: (a) \(\tfrac{2}{7}\).

Example 2 — Conditional Probability (2011-I)

Q: \(A\) and \(B\) are events. \(P(A \cup B) = 0.5\), \(P(B) = 0.8\), \(P(A \mid B) = 0.4\). Find \(P(A \cap B)\).

• Use definition: $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$$
• $$P(A \cap B) = P(A \mid B) \times P(B) = 0.4 \times 0.8 = 0.32.$$
• Check via addition theorem: $$P(A \cup B) = P(A) + P(B) - P(A \cap B) \implies 0.5 = P(A) + 0.8 - 0.32 \implies P(A) = 0.02.$$
• Answer: \(P(A \cap B) = 0.32\).

Example 3 — Binomial Distribution Mode (2013-I)

Q: \(X\) follows a binomial distribution with \(n = 100\) and \(p = \tfrac{1}{3}\). For which value of \(r\) is \(P(X = r)\) maximum?

• The mode of \(B(n, p)\) is found using \((n+1)p\) when this is not an integer.
• $$(n+1)p = 101 \times \tfrac{1}{3} = \tfrac{101}{3} \approx 33.67 \quad \text{— not an integer.}$$
• $$\text{Mode} = \lfloor 33.67 \rfloor = 33.$$
• Answer: \(r = 33\).

Example 4 — Independent Events (2016-I)

Q: Two independent events \(A\) and \(B\) have \(P(A) = \tfrac{1}{3}\) and \(P(B) = \tfrac{1}{4}\). What is the probability that exactly one of the two events occurs?

• $$P(\text{exactly one}) = P(A \cap B') + P(A' \cap B).$$
• Since \(A\) and \(B\) are independent: $$P(A \cap B') = P(A) \cdot P(B') = \tfrac{1}{3} \cdot \tfrac{3}{4} = \tfrac{3}{12} = \tfrac{1}{4}.$$
• $$P(A' \cap B) = P(A') \cdot P(B) = \tfrac{2}{3} \cdot \tfrac{1}{4} = \tfrac{2}{12} = \tfrac{1}{6}.$$
• $$P(\text{exactly one}) = \tfrac{1}{4} + \tfrac{1}{6} = \tfrac{3}{12} + \tfrac{2}{12} = \tfrac{5}{12}.$$

Example 5 — Binomial Mean and Variance (2015-I)

Q: Mean and variance of a binomial variate \(X\) are 2 and 1 respectively. Find \(P(X = 0)\).

• Mean \(= np = 2\) and variance \(= npq = 1\).
• Divide: $$\frac{npq}{np} = q = \tfrac{1}{2}, \quad \text{so } p = \tfrac{1}{2}.$$
• From \(np = 2\) and \(p = \tfrac{1}{2}\): \(n = 4\).
• $$P(X = 0) = \binom{4}{0} \left(\tfrac{1}{2}\right)^{0} \left(\tfrac{1}{2}\right)^{4} = 1 \cdot 1 \cdot \tfrac{1}{16} = \tfrac{1}{16}.$$
• Answer: \(P(X = 0) = \tfrac{1}{16}\).

Example 6 — Conditional Probability on Two Dice

Q: A pair of dice is thrown. Given that the sum is an even number, find the probability that the sum is 8.

• Event \(B\) (given): sum is even. Even sums (2, 4, 6, 8, 10, 12) have outcome counts \(1 + 3 + 5 + 5 + 3 + 1 = 18\), so \(n(B) = 18\).
• Event \(A\): sum is 8. Outcomes are \((2,6), (3,5), (4,4), (5,3), (6,2)\) — all already even, so \(A \subset B\) and \(n(A \cap B) = 5\).
• Apply conditional probability: $$P(A \mid B) = \frac{n(A \cap B)}{n(B)} = \frac{5}{18}.$$
• Answer: \(\tfrac{5}{18}\). Note: the unconditional \(P(\text{sum} = 8) = \tfrac{5}{36}\) — conditioning on "even" doubles it.

Example 7 — Binomial: Exactly 3 Heads in 5 Tosses

Q: A fair coin is tossed 5 times. Find the probability of getting exactly 3 heads.

• This is binomial with \(n = 5\), \(p = \tfrac{1}{2}\) (head), \(q = \tfrac{1}{2}\) (tail), \(r = 3\).
• Apply the formula: $$P(X = 3) = \binom{5}{3} \left(\tfrac{1}{2}\right)^{3} \left(\tfrac{1}{2}\right)^{2}.$$
• Compute \(\binom{5}{3} = 10\) and \(\left(\tfrac{1}{2}\right)^{5} = \tfrac{1}{32}\), so $$P(X = 3) = 10 \times \tfrac{1}{32} = \tfrac{5}{16}.$$
• Answer: 5/16.

Exam Shortcuts (Pro-Tips)

Probability has the highest weightage in NDA Mathematics, so setters love planting time-sink questions. The five hacks below collapse 2–3 minutes of calculation into 10 seconds each. Memorise them — every shortcut here has appeared in a previous NDA paper.

Shortcut 1 — Dice Sum Triangle

When two fair dice are thrown, the number of outcomes for each possible sum follows a triangular pattern. For sums from 2 to 7, the count is simply sum − 1; sums 8 to 12 mirror back down.

Example: \(P(\text{sum} = 6) = \tfrac{5}{36}\) directly, no enumeration needed.

Shortcut 2 — The "53 Sundays" Leap Year Rule

NDA has tested this exact problem in 2010-I, 2012-I, and 2013-II. A normal year has 365 days = 52 weeks + 1 extra day; a leap year has 52 weeks + 2 extra consecutive days. The extra day(s) decide whether a specific weekday occurs 53 times.

Shortcut 3 — "At Least One" Always Means Complement

Any question asking the probability that "at least one" of several independent events occurs is solved by the complement — not by enumerating combinations. For independent events \(A, B, C\) with failure probabilities \(\bar{A}, \bar{B}, \bar{C}\):

This is the "target destroyed" pattern from NDA 2015-I and 2016-II — whenever two or more people each hit a target with given probability, this is the formula.

Shortcut 4 — One-of-Each-Suit Card Draw

Drawing 4 cards from a 52-card deck and asking the probability of one card from each suit is a recurring NDA pattern. Skip the multi-step multiplication; use the direct ratio:

Shortcut 5 — Binomial Mode in One Step

For a binomial distribution \(B(n, p)\), the most probable value of \(X\) is given directly by \((n+1)p\) — no need to compute \(P(X = r)\) for every \(r\) and compare.

NDA 2013-I: \(n = 100, p = \tfrac{1}{3}\) gives \((n+1)p = \tfrac{101}{3} \approx 33.67\), so mode = 33. Solved in 5 seconds.

Common Question Patterns

Pattern 1 — Ball Drawing (with and without replacement)

A box has \(m\) white and \(n\) black balls. Two balls are drawn. Without replacement, treat the draws as dependent events: $$P(\text{both black}) = \frac{n}{m+n} \times \frac{n-1}{m+n-1}.$$ With replacement, the two draws are independent. NDA 2014-I gave: 3 white and 2 black balls, drawn without replacement — \(P(\text{both black}) = \tfrac{2}{5} \times \tfrac{1}{4} = \tfrac{2}{20} = \tfrac{1}{10}\).

Pattern 2 — Letter Arrangement

Find the probability that two specific letters do NOT come together in a random arrangement. For 'UNIVERSITY' (10 letters, 2 I's): total arrangements \(= \tfrac{10!}{2!}\); arrangements where both I's are together \(= 9!\) (treat II as one unit). \(P(\text{together}) = \tfrac{9!}{10!/2!} = \tfrac{2!}{10} = \tfrac{1}{5}\). So \(P(\text{not together}) = 1 - \tfrac{1}{5} = \tfrac{4}{5}\). This appeared in both 2011-II and 2014-II.

Pattern 3 — Geometric Probability

A point is chosen at random inside a region. \(P(\text{event}) = \tfrac{\text{favourable area}}{\text{total area}}\). NDA 2016-I: rectangle \(6 \times 5\) inches, point at least 1 inch from every edge. The inner rectangle is \(4 \times 3 = 12\) sq in. Total area \(= 30\) sq in. \(P = \tfrac{12}{30} = \tfrac{2}{5}\).

Pattern 4 — Dice Sum Problems

For two dice (\(n(S) = 36\)), the sum 7 has the most outcomes (6 pairs), giving \(P = \tfrac{1}{6}\). For sum \(\ge 4\) with two dice (NDA 2016-I), it is easier to find \(P(\text{sum} \le 3) = P(2) + P(3) = \tfrac{1}{36} + \tfrac{2}{36} = \tfrac{3}{36} = \tfrac{1}{12}\), so \(P(\text{sum} \ge 4) = 1 - \tfrac{1}{12} = \tfrac{11}{12}\). Wait — NDA gave answer choices including \(\tfrac{11}{12}\) for this question.

Pattern 5 — Statements About Event Types

NDA frequently asks you to classify or verify statements about mutually exclusive, independent, exhaustive, and elementary events. Anchor these definitions. An elementary event has exactly one sample point (NDA 2012-II tested this directly). Exhaustive events have union = sample space. Mutually exclusive means their intersection is empty.

Preparation Strategy

Probability is one of those topics where consistent practice beats long study sessions. Here is a structured plan that works for NDA timelines:

Week 1 — Foundations. Start with sample space and classical probability. Drill 10–15 problems involving coins, dice, and cards until you can set up \(n(S)\) and \(n(E)\) in under 20 seconds. Then move to the addition theorem — practice problems that give \(P(A)\), \(P(B)\), and \(P(A \cup B)\) and ask for \(P(A \cap B)\), or vice versa. Connect this chapter to Permutations and Combinations — every counting-based probability problem uses combinations.

Week 2 — Conditional Probability and Independence. The definition \(P(A \mid B) = \tfrac{P(A \cap B)}{P(B)}\) is the most tested relationship in the chapter. Practice 10 problems where you have to identify which two of \(P(A), P(B), P(A \cap B), P(A \mid B), P(A \cup B)\) are given and which to find. Also master: if \(A\) and \(B\) are independent, then \(A'\) and \(B\), \(A\) and \(B'\), \(A'\) and \(B'\) are all independent pairs.

Week 3 — Bayes' Theorem and Binomial. Bayes' theorem problems always follow the same structure: two or more "groups" with known base rates, a common outcome, reverse-find the group. Tabulate: \(P(\text{group}), P(\text{outcome} \mid \text{group})\), then apply the formula. For binomial, fix the formula \(P(X = r) = \binom{n}{r} p^{r} q^{n-r}\) and drill mean \(= np\), variance \(= npq\) problems.

Mock tests. After three weeks of concept work, shift to timed practice. Take NDA mock tests and track which probability subtopic causes the most errors. Most students lose marks on conditional probability (confusing P(A | B) with P(B | A)) and on Bayes' theorem setup. Fix those specifically.

Calendar and arrangement problems. These are gift questions — they recur almost verbatim. Memorise: leap year has 2 extra days; non-leap year has 1 extra day; \(P(\text{53rd occurrence of a weekday in a leap year}) = \tfrac{2}{7}\); \(P(\text{in a non-leap year}) = \tfrac{1}{7}\) for most days. For letter arrangement problems, always compute \(P(\text{together})\) first using the "glue" method, then subtract from 1.

Also review Statistics to understand how probability distributions connect to descriptive statistics. The Sets, Relations and Functions chapter underpins the formal logic of event operations — a 30-minute review there will sharpen your thinking on mutually exclusive and exhaustive events.

Frequently Asked Questions

What is the difference between mutually exclusive and independent events?

Mutually exclusive events cannot occur simultaneously — their intersection is empty, so \(P(A \cap B) = 0\). Independent events can both occur; it just means the occurrence of one does not affect the probability of the other, so \(P(A \cap B) = P(A) \cdot P(B)\). Two non-trivial events cannot be both mutually exclusive and independent at the same time. NDA tests this distinction frequently — NDA 2011-I explicitly asked for \(P(A \cap B)\) given \(A\) and \(B\) are mutually exclusive.

How do I quickly solve calendar probability problems?

A leap year has 366 days = 52 weeks + 2 extra days. A non-leap year has 365 days = 52 weeks + 1 extra day. For 53 occurrences of a given weekday in a leap year, the 2 extra days must include that weekday — there are 2 such consecutive-day pairs out of 7, giving \(P = \tfrac{2}{7}\). For a non-leap year, only 1 extra day exists, giving \(P = \tfrac{1}{7}\). December always has 31 days = 4 weeks + 3 extra days; for 5 Sundays, 3 specific pairs out of 7 work, giving \(P = \tfrac{3}{7}\).

What is the formula for the mode of a binomial distribution?

For \(B(n, p)\), the mode (most probable value) is the integer part of \((n+1)p\) when \((n+1)p\) is not an integer. If \((n+1)p\) is exactly an integer, there are two modes: \((n+1)p\) and \((n+1)p - 1\). NDA 2013-I asked this with \(n = 100, p = \tfrac{1}{3}\): \((n+1)p = \tfrac{101}{3} \approx 33.67\), so mode = 33.

How do I find P(X = 0) given mean and variance of a binomial distribution?

Mean \(= np\) and variance \(= npq\). Divide variance by mean to get \(q = \tfrac{npq}{np}\). Then \(p = 1 - q\). Use mean \(= np\) to find \(n\). Finally, \(P(X = 0) = q^{n}\). Example from NDA 2015-I: mean = 2, variance = 1 \(\to q = \tfrac{1}{2}, p = \tfrac{1}{2}, n = 4 \to P(X = 0) = \left(\tfrac{1}{2}\right)^{4} = \tfrac{1}{16}\).

What is Bayes' theorem and when is it used in NDA?

Bayes' theorem computes the probability of a cause given an observed effect. NDA typically presents it as: two or more boxes (or groups), each with a known probability of being chosen and a known probability of producing a specific outcome. You observe the outcome and must find which box (group) it likely came from. Apply: $$P(\text{box}_i \mid \text{outcome}) = \frac{P(\text{box}_i) \cdot P(\text{outcome} \mid \text{box}_i)}{\sum_j P(\text{box}_j) \cdot P(\text{outcome} \mid \text{box}_j)}.$$ NDA 2014-II and 2015-II both had Bayes' theorem questions in this format.

How do I handle "exactly one of two events occurs" questions?

\(P(\text{exactly one of } A \text{ or } B) = P(A \cap B') + P(A' \cap B)\). If \(A\) and \(B\) are independent: \(P(A \cap B') = P(A) \cdot (1 - P(B))\) and \(P(A' \cap B) = (1 - P(A)) \cdot P(B)\). Add the two. NDA 2016-I tested this with \(P(A) = \tfrac{1}{3}\) and \(P(B) = \tfrac{1}{4}\) — answer was \(\tfrac{1}{3} \cdot \tfrac{3}{4} + \tfrac{2}{3} \cdot \tfrac{1}{4} = \tfrac{1}{4} + \tfrac{1}{6} = \tfrac{5}{12}\).

Which topics in NDA Maths connect to Probability?

Permutations and Combinations is the most direct link — classical probability problems require counting favourable and total outcomes using combinations. Sets and Logic underpin event operations (union, intersection, complement). Statistics connects through probability distributions. Build your combinatorics base first, then tackle probability. Mock tests at Defence Road cover all three in integrated papers.

Related Topics

NDA Maths Permutations and Combinations NDA Maths Statistics NDA Maths Sets, Relations and Functions NDA Maths Sequences and Series NDA NDA Mathematics — Full Syllabus