NDA Maths · Probability

Bounds on Probability

The inequalities that constrain probabilities — Fréchet and Boole bounds, the min/max of unions and intersections, and the identity-statement traps.

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Why this matters

This is the chapter's advanced corner — only 11 questions, but they are mostly MODERATE/HARD and almost always phrased as "which of the following statements are correct". The marks come from knowing that P(A), P(B) and P(A and B) cannot be chosen freely: the intersection has a forced floor, the union has a ceiling, and several plausible-looking identities are subtly wrong. Learn the four bounds and the exactly-one trap and this section becomes quick, defensible marks rather than guesswork.

Test yourself — a quick 15-question Probability recall quiz, instant score.

Concept 1 of 3

Fréchet and Boole bounds

Intuition

You cannot pick

P(A)

P(B)

and

P(A \cap B)

independently — they constrain each other. The overlap cannot be larger than the smaller event, and when

P(A) + P(B)

exceeds 1 the events are forced to overlap by at least the excess. The union, in turn, can never exceed the sum.

Definition

For any two events (all from $0 \le P \le 1$ and the addition rule):

Intersection (Fréchet): $\max(0,\ P(A) + P(B) - 1) \le P(A \cap B) \le \min(P(A), P(B))$ .
Union: $\max(P(A), P(B)) \le P(A \cup B) \le \min(1,\ P(A) + P(B))$ .
Boole's inequality: the upper union bound $P(A \cup B) \le P(A) + P(B)$ .

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

Worked example

P(A) = 0.7

and

P(B) = 0.6

, find the range of possible values of

P(A \cap B)

Lower bound: $\max(0,\ 0.7 + 0.6 - 1) = \max(0,\ 0.3) = 0.3$ .
Upper bound: $\min(0.7,\ 0.6) = 0.6$ .
So $0.3 \le P(A \cap B) \le 0.6$ .

Answer:

0.3 \le P(A \cap B) \le 0.6

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1ProbabilityEASY

Consider the following relations for two events

E

and

F

: 1.

P(E\cap F)\ge P(E)+P(F)-1

P(E\cup F)=P(E)+P(F)+P(E\cap F)

P(E\cup F)\le P(E)+P(F)

Which of the above relations is/are correct?

[Q101 · Sep · 2021]

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.

Drill 1 more on fréchet and boole bounds

Concept 2 of 3

Minimum and maximum of combined probabilities

Intuition

To find the extreme value of

P(A \cup B)

P(A \cap B)

P(A)+P(B)

, push the overlap to whichever boundary the bounds allow. The identity

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

converts between them.

Definition

The four extreme values (linked by the identity $P(A) + P(B) = P(A \cup B) + P(A \cap B)$ ):

Union, min: $P(A \cup B) = \max(P(A), P(B))$
Union, max: $P(A \cup B) = \min(1,\ P(A)+P(B))$
Intersection, min: $P(A \cap B) = \max(0,\ P(A)+P(B)-1)$
Intersection, max: $P(A \cap B) = \min(P(A), P(B))$

Worked example

P(A) = \dfrac{2}{3}

and

P(B) = \dfrac{3}{5}

, find the minimum value of

P(A \cup B)

and the maximum value of

P(A \cap B)

Both extremes occur at maximum overlap ( $B$ pushed as far inside $A$ as possible).
Minimum union: $\min P(A \cup B) = \max(P(A), P(B)) = \max\left(\dfrac{2}{3}, \dfrac{3}{5}\right) = \dfrac{2}{3}$ .
Maximum intersection: $\max P(A \cap B) = \min(P(A), P(B)) = \min\left(\dfrac{2}{3}, \dfrac{3}{5}\right) = \dfrac{3}{5}$ .

Answer:

\min P(A \cup B) = \dfrac{2}{3}

;

\max P(A \cap B) = \dfrac{3}{5}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2ProbabilityEASY

A

and

B

are two events such that

P(A)=\frac{3}{4}

and

P(B)=\frac{5}{8}

, then consider the following statements: 1. The minimum value of

P(A\cup B)

\frac{3}{4}

. 2. The maximum value of

P(A\cap B)

\frac{5}{8}

. Which of the above statements is/are correct?

[Q120 · Apr · 2021]

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.

Drill 4 more on minimum and maximum of combined probabilities

Concept 3 of 3

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

Intuition

NDA loves the format "consider the following statements" with a list of probability identities, where one is subtly wrong. Test each against the definitions — the planted error is almost always the exactly-one formula or a mis-aimed subtraction.

Definition

Reliable identities:

Only A: $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ (subtract the overlap from the SAME event)
Exactly one: $P(\text{exactly one of } A, B) = P(A) + P(B) - 2P(A \cap B)$
Union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Subset: if $B \subseteq A$ then $P(A \cap \bar{B}) = P(A) - P(B)$
**At least two of $A,B,C$ :** $P(A\cap B)+P(B\cap C)+P(C\cap A)-2P(A\cap B\cap C)$
**Exactly two of $A,B,C$ :** $P(A\cap B)+P(B\cap C)+P(C\cap A)-3P(A\cap B\cap C)$
Planted error: "exactly one" written with a single $-P(A \cap B)$ (that is the union) instead of $-2P(A \cap B)$ ; for three events, the wrong coefficient on $P(A\cap B\cap C)$

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

Worked example

Which of the following are correct? (1)

P(A \cap \bar{B}) = P(A) - P(A \cap B)

. (2)

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

. (3)

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

(1) Correct — $A$ splits into $A \cap B$ and $A \cap \bar{B}$ , so $P(A \cap \bar{B}) = P(A) - P(A \cap B)$ .
(2) Correct — this is the addition rule.
(3) Incorrect — exactly one is $P(A) + P(B) - 2P(A \cap B)$ ; statement (3) is actually the formula for the union.

Answer:Only (1) and (2) are correct.

Practice this conceptself-check · 6 quick reps

From the bank · past-year question

Example 3ProbabilityMODERATE

Which of the following probability statements are correct? 1.

P(\bar{A}\cup B) = P(\bar{A})+P(B)-P(\bar{A}\cap B)

P(A\cap\bar{B}) = P(B)-P(A\cap B)

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

[Q102 · Apr · 2018]

"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

Drill 3 more on identity-statement traps ("which are correct?")

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Fréchet and Boole bounds
Bounds on intersection and union
$\max\big(0,\,P(A)+P(B)-1\big) \le P(A \cap B) \le \min\big(P(A),\,P(B)\big)$
Identity-statement traps ("which are correct?")
Exactly one vs union
$P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B), \qquad P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Watch out for (7)

The intersection floor $P(A)+P(B)-1$ only bites when the sum exceeds 1
→ Fréchet and Boole bounds
The union floor is $\max(P(A),P(B))$ , not $P(A)+P(B)$
→ Fréchet and Boole bounds
Minimum union = max of the two probabilities, maximum union = their sum (capped at 1)
→ Minimum and maximum of combined probabilities
Convert via $P(A)+P(B) = P(A\cup B) + P(A\cap B)$
→ Minimum and maximum of combined probabilities
"Exactly one" subtracts the overlap TWICE
→ Identity-statement traps ("which are correct?")
"At least two of three" subtracts the triple overlap TWICE (−2), not −1 or −3
→ Identity-statement traps ("which are correct?")
$P(A \cap \bar{B}) = P(A) - P(A \cap B)$ , not $P(A) - P(B)$
→ Identity-statement traps ("which are correct?")

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