NDA Mathematics · Formula sheet

Probability formulas

22 formulas and 50 common traps for NDA Mathematics Probability, grouped by subtopic.

Classical Probability & Counting

Classical (theoretical) probability

Classical probability

P(E) = \dfrac{n(E)}{n(S)} = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}

$n(E)$ number of outcomes that make $E$ occur
$n(S)$ total number of equally likely outcomes

Geometric probability (length / area / volume ratio)

Geometric probability

P(E) = \dfrac{\text{favourable measure}}{\text{total measure}}\quad(\text{length / area / volume})

Axioms, range, complement, and odds

Complement rule and range

0 \le P(E) \le 1, \qquad P(S)=1, \qquad P(E') = 1 - P(E)

$E'$ complement of $E$ — the event that $E$ does not occur
$P(\varnothing)$ probability of the impossible event, equal to $0$

Selection probability with combinations

Selection probability (combinations)

P = \dfrac{\dbinom{a}{k}\dbinom{b}{r-k}}{\dbinom{a+b}{r}}

$a, b$ counts of the two types of object
$k$ how many of the first type the event requires
$r$ total number drawn

Probability with dice

Two-dice sample space

n(S) = 6^2 = 36, \qquad P(\text{sum}=7) = \dfrac{6}{36} = \dfrac{1}{6}

$(a,b)$ ordered pair: $a$ on the first die, $b$ on the second

Probability with coins

Coin tosses

n(S)=2^{n}, \qquad P(\text{at least one head in } n) = 1 - \left(\tfrac{1}{2}\right)^{n}

$n$ number of tosses
$p$ probability of a head on one toss ( $\tfrac{1}{2}$ if fair)

Probability with arrangements

Arrangement probability (two together)

P(\text{two specified together}) = \dfrac{2\,(n-1)!}{n!} = \dfrac{2}{n}

$n$ number of objects being arranged
$2\,(n-1)!$ favourable: glue the pair ( $(n-1)!$ ) and order them internally ( $2!$ )

Choosing numbers with a property

Counting favourable numbers

P = \dfrac{\#\{x : x \text{ has the property}\}}{n}

$\lfloor n/d\rfloor$ how many of $1,\dots,n$ are multiples of $d$

Common traps

An event is a SET of outcomes, not a single outcome

"Getting a number greater than 4" on a die is the event

\{5,6\}

— two outcomes — not one. Mis-reading an event as a single outcome is the most common source of a wrong favourable-count.

Write or size the sample space before you count anything

Almost every classical-probability error is a counting error in

n(S)

n(E)

. Fix

n(S)

first (it is

2^n

for coins,

6^k

for dice,

\binom{n}{r}

for unordered selections), then count favourable outcomes against it.

The classical formula needs EQUALLY LIKELY outcomes

P=n(E)/n(S)

is only valid when every outcome is equally likely. A loaded die, a biased coin, or faces repeated unevenly (a die with two 4s) breaks the assumption — weight each outcome by its own probability instead of counting.

Favourable is a subset of total, so $P$ can never exceed 1

If a computation gives

P>1

P<0

, the favourable or total count is wrong. Recount before trusting the answer.

Use the right MEASURE — "closer to the centre" is an AREA ratio, not a radius ratio

Distance from the centre

< r/2

is a DISC of radius

r/2

, whose area is

\pi(r/2)^2

. The probability is the area ratio

(r/2)^2/r^2 = \tfrac14

, NOT the radius ratio

\tfrac12

. Matching the measure to the dimension is the whole trap.

Odds are not probability: $a:b$ in favour means $\dfrac{a}{a+b}$ , not $\dfrac{a}{b}$

Convert odds to a probability by putting the favourable parts over the TOTAL parts. Odds in favour

3:2

\tfrac{3}{5}

; odds against

3:2

\tfrac{2}{5}

"At least one…" almost always means use the complement

P(\text{at least one}) = 1 - P(\text{none})

. Counting the cases directly (exactly one, exactly two, …) is slower and a frequent source of arithmetic slips.

"Drawn together / selected at random" means order does NOT matter — use $\binom{n}{r}$ , not permutations

If you accidentally use

{}^{n}P_r

(ordered) in both numerator and denominator the ratio often still works, but mixing one ordered count with one unordered count gives the wrong answer. Keep both counts unordered.

"At least one of a type" is fastest via the complement

P(\text{at least one red}) = 1 - P(\text{no red}) = 1 - \dfrac{\binom{b}{r}}{\binom{a+b}{r}}

. Summing exactly-one, exactly-two, … wastes time.

Two-dice outcomes are ORDERED pairs: $(2,3)$ and $(3,2)$ are different

The sample space has

36

outcomes, not

21

. Counting unordered pairs undercounts every non-doublet event by a factor of two.

Loaded or non-standard dice: faces are NOT equally likely

If a die has two faces showing 4, or even faces are twice as likely as odd, you cannot use

n(E)/36

. Assign each face its probability first, then add up the favourable outcomes' probabilities.

Sequences are ordered: $HT \ne TH$

Count the

2^n

ordered sequences, not the

n+1

"number of heads" buckets — those buckets are not equally likely ("exactly 1 head in 2 tosses" has 2 sequences, "2 heads" has 1).

Biased coin: do not use $2^n$ equally-likely counting

When

P(H) \ne \tfrac{1}{2}

, each sequence has its own probability

p^{h}(1-p)^{t}

. Counting outcomes and dividing by

2^n

is only valid for a fair coin.

"Together" = glue into a block, then multiply by the block's internal arrangements

Forgetting the

2!

for the order within the pair halves the favourable count. For a block of

k

people the internal factor is

k!

Repeated letters divide the total by the factorial of each repeat count

The arrangements of TIRUPATI use

\dfrac{8!}{2!\,2!}

(two T's, two I's), not

8!

. Forgetting this inflates both counts unequally and breaks the ratio.

"Or" on number properties needs inclusion-exclusion

P(\text{mult of 3 or 5}) = P(3) + P(5) - P(15)

. Forgetting to subtract the overlap (multiples of both) double-counts and overstates the probability.

Several numbers chosen at once: denominator is $\binom{n}{r}$ , not $n$

"Three numbers chosen from 1 to 10" has

\binom{10}{3}=120

outcomes. Only use

n

in the denominator when exactly one number is picked.

Event Algebra & the Addition Rule

Learn this subtopic in the notes

The addition rule (inclusion-exclusion)

Addition rule

P(A \cup B) = P(A) + P(B) - P(A \cap B)

$P(A \cap B)$ probability both occur — subtracted to undo double-counting

"Neither" and the complement of a union

Complement of a union (De Morgan)

P(A' \cap B') = 1 - P(A \cup B) = 1 - P(A) - P(B) + P(A \cap B)

$A' \cap B'$ the "neither" region — outside both circles

Mutually exclusive (disjoint) events

Addition rule for mutually exclusive events

A \cap B = \varnothing \;\Rightarrow\; P(A \cup B) = P(A) + P(B)

$\varnothing$ the impossible event — the two cannot co-occur

Exhaustive events (and probabilities that sum to 1)

Mutually exclusive AND exhaustive

A_1 \cup \dots \cup A_n = S \;\text{ and disjoint} \;\Rightarrow\; \sum_{i} P(A_i) = 1

partitionmutually exclusive + exhaustive: exactly one event occurs

Common traps

$A \cup B$ is INCLUSIVE "or" — it contains the overlap

"A or B" in probability always means "A, or B, or both". The exclusive "exactly one" is a different event,

(A \cap B') \cup (A' \cap B)

Read "and" as intersection, "or" as union — do not swap

A \cap B

is the smaller event (both must hold);

A \cup B

is the larger (one suffices). Mixing them up flips the whole calculation.

Do not forget to subtract $P(A \cap B)$

P(A) + P(B)

alone overcounts the overlap. The only time you may skip the subtraction is when

A

and

B

are mutually exclusive (overlap

= 0

Three events need the full inclusion-exclusion, not just three single terms

Subtract all three pairwise intersections, then add back the triple intersection. Stopping after the singles or after the pairs gives a wrong answer.

"Neither" is $1 - P(A \cup B)$ , not $1 - P(A) - P(B)$

Skipping the

+P(A \cap B)

term double-subtracts the overlap and understates the answer. Compute the union correctly first, then take the complement.

De Morgan flips the operation: complement of a union is an intersection

(A \cup B)' = A' \cap B'

(neither), while

(A \cap B)' = A' \cup B'

(not both). Using the wrong one swaps two different events.

Mutually exclusive $\ne$ independent — opposite ideas

Mutually exclusive means

P(A \cap B) = 0

; independent means

P(A \cap B) = P(A)P(B)

. For events with positive probability these cannot hold at once. Questions often test exactly this confusion.

Only drop the overlap term when you are TOLD the events are mutually exclusive

If a problem does not state disjointness (or give

P(A \cap B) = 0

), you must keep the

-P(A \cap B)

term in the addition rule.

Exhaustive alone does not give sum $= 1$

If the events overlap,

\sum P(A_i)

exceeds 1 by the overlaps. The probabilities add to exactly 1 only when the events are BOTH exhaustive AND mutually exclusive (a partition).

Turn a chained ratio into one variable before summing

2P(A) = 3P(B) = 4P(C) = k

means

P(A) = \tfrac{k}{2}

P(B) = \tfrac{k}{3}

P(C) = \tfrac{k}{4}

. Set the sum to 1 to find

k

; a common slip is reading the ratio as

P(A):P(B):P(C) = 2:3:4

(it is actually the reciprocals).

Independent Events & the Multiplication Rule

Learn this subtopic in the notes

Independence and the multiplication rule

Multiplication rule (independent events)

P(A \cap B) = P(A)\,P(B)

$P(A \cap B)$ probability both occur — a product, only when independent

"At least one" via the complement

At least one (independent trials)

P(\text{at least one}) = 1 - \prod_{i}\big(1 - P(A_i)\big)

$1 - P(A_i)$ probability trial $i$ fails
$\prod$ product over all trials — probability all fail

The "problem solved by students" archetype

Problem solved by at least one solver

P(\text{solved}) = 1 - \prod_{i}(1 - p_i)

$p_i$ probability solver $i$ solves it, independently

Finding an unknown probability using independence

Union of independent events

P(A \cup B) = P(A) + P(B) - P(A)P(B) = 1 - P(A')\,P(B')

$P(A')P(B')$ probability neither occurs (independent complements)

Common traps

Independent $\ne$ mutually exclusive

Independent means

P(A \cap B) = P(A)P(B)

(both can happen, just unrelated); mutually exclusive means

P(A \cap B) = 0

(both cannot happen). They are opposite conditions for positive-probability events.

Only multiply when independence is given or physically clear

Drawing without replacement, or events from the same experiment, are usually NOT independent — use conditional probability there. Multiply only when the trials genuinely don't influence each other.

"At least one" = 1 - (all fail), not the sum of individual probabilities

Adding

P(A) + P(B)

double-counts the both-succeed case and can exceed 1. Always go through the complement of "none".

Multiply the FAILURE probabilities, not the success ones, for "none"

"None occurs" means every trial fails, so multiply

(1-p_i)

. A common slip is multiplying the

p_i

(that is "all succeed").

"Problem solved" means at least one solver — use the complement

Do not add the solvers' probabilities (that overcounts and can exceed 1). Compute

1 - \prod(1-p_i)

"Exactly one solves" is a different computation

For exactly one, sum the cases where one succeeds and the rest fail:

p_1(1-p_2) + (1-p_1)p_2

for two solvers. Don't confuse it with "at least one".

For independent events the union is NOT $P(A) + P(B)$

You must subtract

P(A)P(B)

for the overlap.

P(A) + P(B)

(no subtraction) is the mutually-exclusive formula — the wrong model for independent events.

Use $1 - P(A')P(B')$ as a fast check

Computing the union as

1

minus "neither" often avoids a sign slip; both routes must agree.

Conditional Probability, Total Probability & Bayes'

Learn this subtopic in the notes

Conditional probability

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \qquad P(B) > 0

$P(A \cap B)$ probability both occur
$P(B)$ probability of the condition — the new "total"

Multiplication rule & restricted sample space

Multiplication rule / restricted counting

P(A \cap B) = P(B)\,P(A \mid B), \qquad P(A \mid B) = \dfrac{n(A \cap B)}{n(B)}

$n(B)$ number of outcomes in the condition — the restricted total
$n(A \cap B)$ favourable outcomes within the condition

Total probability (over a partition)

Total probability

P(A) = \sum_{i=1}^{n} P(B_i)\,P(A \mid B_i)

$B_i$ the mutually exclusive, exhaustive routes (partition)
$P(A \mid B_i)$ probability of $A$ along route $i$

Bayes' theorem (reversing the conditional)

Bayes' theorem

P(B_k \mid A) = \dfrac{P(B_k)\,P(A \mid B_k)}{\displaystyle\sum_{i} P(B_i)\,P(A \mid B_i)}

numeratorthe chosen route's forward contribution $P(B_k)P(A\mid B_k)$
denominatortotal probability of $A$ over all routes

Common traps

Mind which event is the condition: $P(A\mid B) \ne P(B\mid A)$ in general

The denominator is the probability of the GIVEN event.

P(A\mid B)

divides by

P(B)

;

P(B\mid A)

divides by

P(A)

. They are equal only when

P(A) = P(B)

The condition must have positive probability

P(A\mid B)

is undefined when

P(B) = 0

. Check the condition is possible before dividing.

Under a condition, the TOTAL changes to $n(B)$ , not 36 (or 6)

"Given that …" shrinks the sample space. Divide by the number of outcomes in the condition, not by the original total. Forgetting to restrict is the classic conditional-probability error.

The general multiplication rule needs $P(A \mid B)$ , not $P(A)$

P(A \cap B) = P(A)P(B)

is only the independent case. In general use

P(A \cap B) = P(B)P(A \mid B)

Weight each route by its own probability

P(A)

is not the average of the conditional probabilities unless the routes are equally likely. Always multiply each

P(A \mid B_i)

P(B_i)

before summing.

The routes must be a partition

Total probability requires the

B_i

to be mutually exclusive and to cover every possibility. Missing a route (or overlapping routes) breaks the sum.

Do not confuse $P(B_k \mid A)$ with $P(A \mid B_k)$

The question gives you the forward conditionals

P(A \mid B_i)

(defect rate per machine) and asks for the reverse

P(B_k \mid A)

(which machine, given a defect). Bayes is exactly the tool that flips them — don't report the forward number.

The denominator is the FULL total probability, not just $P(A \mid B_k)$

Divide route

k

's contribution by the sum over ALL routes. Using only the chosen route's term gives 1 every time — a sure sign the denominator is wrong.

Bounds on Probability

Learn this subtopic in the notes

Fréchet and Boole bounds

Bounds on intersection and union

\max\big(0,\,P(A)+P(B)-1\big) \le P(A \cap B) \le \min\big(P(A),\,P(B)\big)

lower bound $P(A)+P(B)-1$ — the forced overlap when the sum exceeds 1 (else 0)
upper bound $\min(P(A),P(B))$ — the overlap can't exceed the smaller event

Identity-statement traps ("which are correct?")

Exactly one vs union

P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B), \qquad P(A \cup B) = P(A) + P(B) - P(A \cap B)

the $-2$ exactly-one excludes the overlap TWICE; the union keeps it once

Common traps

The intersection floor $P(A)+P(B)-1$ only bites when the sum exceeds 1

P(A) + P(B) \le 1

the lower bound is just 0 (the events can be disjoint). Always take

\max(0,\ P(A)+P(B)-1)

— never report a negative lower bound.

The union floor is $\max(P(A),P(B))$ , not $P(A)+P(B)$

P(A) + P(B)

is the union's CEILING (Boole), reached only when the events are disjoint. The smallest the union can be is the larger single probability, reached when one event sits inside the other.

Minimum union = max of the two probabilities, maximum union = their sum (capped at 1)

Students often swap these. The union is SMALLEST when overlap is largest (one event inside the other) and LARGEST when overlap is smallest (disjoint).

Convert via $P(A)+P(B) = P(A\cup B) + P(A\cap B)$

When a question constrains the union and intersection and asks for

P(A)+P(B)

(or vice-versa), this identity is the bridge — minimise/maximise the two right-hand terms independently within their allowed ranges.

"Exactly one" subtracts the overlap TWICE

P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B)

. Writing a single

-P(A \cap B)

gives the union — the most common planted error in these statement questions.

"At least two of three" subtracts the triple overlap TWICE (−2), not −1 or −3

P(\text{at least two of }A,B,C)=\sum P(\text{pairwise})-2P(A\cap B\cap C)

. The pairwise sum counts the all-three region three times; you want it once, so remove it twice. ("Exactly two" removes it three times →

-3

; the NDA option list dangles

-1,-2,-3

to catch this.)

$P(A \cap \bar{B}) = P(A) - P(A \cap B)$ , not $P(A) - P(B)$

Subtract the overlap from the SAME event you are restricting.

P(A) - P(B)

is only valid in the special case

B \subseteq A

Probability formulas

Classical Probability & Counting

Classical (theoretical) probability

Geometric probability (length / area / volume ratio)

Axioms, range, complement, and odds

Selection probability with combinations

Probability with dice

Probability with coins

Probability with arrangements

Choosing numbers with a property

Common traps

Event Algebra & the Addition Rule

The addition rule (inclusion-exclusion)

"Neither" and the complement of a union

Mutually exclusive (disjoint) events

Exhaustive events (and probabilities that sum to 1)

Common traps

Independent Events & the Multiplication Rule

Independence and the multiplication rule

"At least one" via the complement

The "problem solved by students" archetype

Finding an unknown probability using independence

Common traps

Conditional Probability, Total Probability & Bayes'

Multiplication rule & restricted sample space

Total probability (over a partition)

Bayes' theorem (reversing the conditional)

Common traps

Bounds on Probability

Fréchet and Boole bounds

Identity-statement traps ("which are correct?")

Common traps

More NDA Mathematics formula sheets