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.

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.

Concept 1 of 3

Fréchet and Boole bounds

Intuition

You cannot pick P(A)P(A), P(B)P(B) and P(AB)P(A \cap B) independently — they constrain each other. The overlap cannot be larger than the smaller event, and when P(A)+P(B)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 0P10 \le P \le 1 and the addition rule):

  • Intersection (Fréchet): max(0, P(A)+P(B)1)P(AB)min(P(A),P(B))\max(0,\ P(A) + P(B) - 1) \le P(A \cap B) \le \min(P(A), P(B)).
  • Union: max(P(A),P(B))P(AB)min(1, P(A)+P(B))\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(AB)P(A)+P(B)P(A \cup B) \le P(A) + P(B).

Bounds on intersection and union

max(0,P(A)+P(B)1)P(AB)min(P(A),P(B))\max\big(0,\,P(A)+P(B)-1\big) \le P(A \cap B) \le \min\big(P(A),\,P(B)\big)
  • lower boundP(A)+P(B)1P(A)+P(B)-1 — the forced overlap when the sum exceeds 1 (else 0)
  • upper boundmin(P(A),P(B))\min(P(A),P(B)) — the overlap can't exceed the smaller event

Worked example

If P(A)=0.7P(A) = 0.7 and P(B)=0.6P(B) = 0.6, find the range of possible values of P(AB)P(A \cap B).
  1. Lower bound: max(0, 0.7+0.61)=max(0, 0.3)=0.3\max(0,\ 0.7 + 0.6 - 1) = \max(0,\ 0.3) = 0.3.
  2. Upper bound: min(0.7, 0.6)=0.6\min(0.7,\ 0.6) = 0.6.
  3. So 0.3P(AB)0.60.3 \le P(A \cap B) \le 0.6.
Answer:0.3P(AB)0.60.3 \le P(A \cap B) \le 0.6
Practice this conceptself-check · 4 quick reps

Try it yourself

If P(A)=0.5P(A) = 0.5 and P(B)=0.4P(B) = 0.4, find the range of possible values of P(AB)P(A \cup B).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    P(A)=0.8,P(B)=0.5P(A)=0.8, P(B)=0.5. Minimum P(AB)P(A\cap B)?
  2. 2.
    P(A)=0.8,P(B)=0.5P(A)=0.8, P(B)=0.5. Maximum P(AB)P(A\cap B)?
  3. 3.
    P(A)=0.3,P(B)=0.4P(A)=0.3, P(B)=0.4. Minimum P(AB)P(A\cap B)?
  4. 4.
    P(A)=0.6,P(B)=0.7P(A)=0.6,P(B)=0.7. Maximum P(AB)P(A\cup B)?

From the bank · past-year question

Example 1ProbabilityEASY
Consider the following relations for two events EE and FF: 1. P(EF)P(E)+P(F)1P(E\cap F)\ge P(E)+P(F)-1 2. P(EF)=P(E)+P(F)+P(EF)P(E\cup F)=P(E)+P(F)+P(E\cap F) 3. P(EF)P(E)+P(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)1P(A)+P(B)-1 only bites when the sum exceeds 1

If P(A)+P(B)1P(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)\max(0,\ P(A)+P(B)-1) — never report a negative lower bound.

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

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.

Concept 2 of 3

Minimum and maximum of combined probabilities

Intuition

To find the extreme value of P(AB)P(A \cup B), P(AB)P(A \cap B) or P(A)+P(B)P(A)+P(B), push the overlap to whichever boundary the bounds allow. The identity P(A)+P(B)=P(AB)+P(AB)P(A) + P(B) = P(A \cup B) + P(A \cap B) converts between them.

Definition

Minimum P(AB)=max(P(A),P(B))P(A \cup B) = \max(P(A), P(B)); maximum P(AB)=min(1, P(A)+P(B))P(A \cup B) = \min(1,\ P(A)+P(B)). Minimum P(AB)=max(0, P(A)+P(B)1)P(A \cap B) = \max(0,\ P(A)+P(B)-1); maximum P(AB)=min(P(A),P(B))P(A \cap B) = \min(P(A), P(B)). Because P(A)+P(B)=P(AB)+P(AB)P(A) + P(B) = P(A \cup B) + P(A \cap B), a constraint on the union and intersection bounds their sum directly.

Linking identity

P(A)+P(B)=P(AB)+P(AB)P(A) + P(B) = P(A \cup B) + P(A \cap B)
  • useto bound P(A)+P(B)P(A)+P(B) from bounds on the union and intersection (and vice-versa)

Worked example

If P(A)=23P(A) = \dfrac{2}{3} and P(B)=35P(B) = \dfrac{3}{5}, find the minimum value of P(AB)P(A \cup B) and the maximum value of P(AB)P(A \cap B).
  1. Both extremes occur at maximum overlap (BB pushed as far inside AA as possible).
  2. Minimum union: minP(AB)=max(P(A),P(B))=max(23,35)=23\min P(A \cup B) = \max(P(A), P(B)) = \max\left(\dfrac{2}{3}, \dfrac{3}{5}\right) = \dfrac{2}{3}.
  3. Maximum intersection: maxP(AB)=min(P(A),P(B))=min(23,35)=35\max P(A \cap B) = \min(P(A), P(B)) = \min\left(\dfrac{2}{3}, \dfrac{3}{5}\right) = \dfrac{3}{5}.
Answer:minP(AB)=23\min P(A \cup B) = \dfrac{2}{3}; maxP(AB)=35\max P(A \cap B) = \dfrac{3}{5}.
Practice this conceptself-check · 4 quick reps

Try it yourself

Two events satisfy P(AB)0.75P(A \cup B) \ge 0.75 and 0.125P(AB)0.3750.125 \le P(A \cap B) \le 0.375. What is the minimum value of P(A)+P(B)P(A) + P(B)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    P(A)=0.6,P(B)=0.5P(A)=0.6,P(B)=0.5. Minimum P(AB)P(A\cup B)?
  2. 2.
    P(A)=0.6,P(B)=0.5P(A)=0.6,P(B)=0.5. Maximum P(AB)P(A\cup B)?
  3. 3.
    P(AB)=0.8,P(AB)=0.3P(A\cup B)=0.8, P(A\cap B)=0.3. P(A)+P(B)P(A)+P(B)?
  4. 4.
    P(A)=12,P(B)=12P(A)=\tfrac{1}{2},P(B)=\tfrac{1}{2}. Minimum P(AB)P(A\cap B)?

From the bank · past-year question

Example 2ProbabilityEASY
If AA and BB are two events such that P(A)=34P(A)=\frac{3}{4} and P(B)=58P(B)=\frac{5}{8}, then consider the following statements: 1. The minimum value of P(AB)P(A\cup B) is 34\frac{3}{4}. 2. The maximum value of P(AB)P(A\cap B) is 58\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(AB)+P(AB)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)P(A)+P(B) (or vice-versa), this identity is the bridge — minimise/maximise the two right-hand terms independently within their allowed ranges.

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: P(ABˉ)=P(A)P(AB)P(A \cap \bar{B}) = P(A) - P(A \cap B) (subtract the overlap from the SAME event); P(exactly one of A,B)=P(A)+P(B)2P(AB)P(\text{exactly one of } A, B) = P(A) + P(B) - 2P(A \cap B); P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B); and if BAB \subseteq A then P(ABˉ)=P(A)P(B)P(A \cap \bar{B}) = P(A) - P(B). The classic planted error writes "exactly one" with a single P(AB)-P(A \cap B) (that is the union) instead of 2P(AB)-2P(A \cap B).

Exactly one vs union

P(exactly one)=P(A)+P(B)2P(AB),P(AB)=P(A)+P(B)P(AB)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 2-2exactly-one excludes the overlap TWICE; the union keeps it once

Worked example

Which of the following are correct? (1) P(ABˉ)=P(A)P(AB)P(A \cap \bar{B}) = P(A) - P(A \cap B). (2) P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B). (3) P(exactly one of A,B)=P(A)+P(B)P(AB)P(\text{exactly one of } A, B) = P(A) + P(B) - P(A \cap B).
  1. (1) Correct — AA splits into ABA \cap B and ABˉA \cap \bar{B}, so P(ABˉ)=P(A)P(AB)P(A \cap \bar{B}) = P(A) - P(A \cap B).
  2. (2) Correct — this is the addition rule.
  3. (3) Incorrect — exactly one is P(A)+P(B)2P(AB)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 · 4 quick reps

Try it yourself

If BAB \subseteq A, simplify P(ABˉ)P(A \cap \bar{B}).

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    P(exactly one of A,B)P(\text{exactly one of } A,B) in terms of P(A),P(B),P(AB)P(A),P(B),P(A\cap B)?
  2. 2.
    P(A)=0.6,P(AB)=0.25P(A)=0.6,P(A\cap B)=0.25. P(ABˉ)P(A\cap\bar{B})?
  3. 3.
    P(A)=0.5,P(B)=0.4,P(AB)=0.2P(A)=0.5,P(B)=0.4,P(A\cap B)=0.2. P(exactly one)P(\text{exactly one})?
  4. 4.
    If BAB\subseteq A, is P(AB)=P(B)P(A\cap B)=P(B)?

From the bank · past-year question

Example 3ProbabilityMODERATE
Which of the following probability statements are correct? 1. P(AˉB)=P(Aˉ)+P(B)P(AˉB)P(\bar{A}\cup B) = P(\bar{A})+P(B)-P(\bar{A}\cap B) 2. P(ABˉ)=P(B)P(AB)P(A\cap\bar{B}) = P(B)-P(A\cap B) 3. P(AB)=P(B)P(AB)P(A\cap B) = P(B)\,P(A\mid B)

[Q102 · Apr · 2018]

"Exactly one" subtracts the overlap TWICE

P(exactly one)=P(A)+P(B)2P(AB)P(\text{exactly one}) = P(A) + P(B) - 2P(A \cap B). Writing a single P(AB)-P(A \cap B) gives the union — the most common planted error in these statement questions.

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

Subtract the overlap from the SAME event you are restricting. P(A)P(B)P(A) - P(B) is only valid in the special case BAB \subseteq A.

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 (3)

  • Fréchet and Boole bounds

    Bounds on intersection and union

    max(0,P(A)+P(B)1)P(AB)min(P(A),P(B))\max\big(0,\,P(A)+P(B)-1\big) \le P(A \cap B) \le \min\big(P(A),\,P(B)\big)
  • Minimum and maximum of combined probabilities

    Linking identity

    P(A)+P(B)=P(AB)+P(AB)P(A) + P(B) = P(A \cup B) + P(A \cap B)
  • Identity-statement traps ("which are correct?")

    Exactly one vs union

    P(exactly one)=P(A)+P(B)2P(AB),P(AB)=P(A)+P(B)P(AB)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 (6)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1ProbabilityMODERATE
Consider the following statements : 1. If AA and BB are mutually exclusive events, then it is possible that P(A)=P(B)=06P(A) = P(B) = 0\cdot6. 2. If AA and BB are any two events such that P(AB)=1P(A|B) = 1, then P(BˉAˉ)=1P(\bar{B}|\bar{A}) = 1. Which of the above statements is/are correct ?

[Q114 · Sep · 2019]

Example 2ProbabilityMODERATE
Let AA and BB be two events such that P(AB)0.75P(A\cup B)\geq0.75 and 0.125P(AB)0.3750.125\leq P(A\cap B)\leq0.375.
What is the minimum value of P(A)+P(B)P(A)+P(B)?

[Q111 · Apr · 2024]

Example 3ProbabilityHARD
Which probability statements are correct? 1. P(A not B)=P(A)-P(B) if B subset A 2. P(A alone or B alone)=P(A)+P(B)-P(A intersect B) 3. P(A union B)=P(A)+P(B) if mutually exclusive

[Q118 · Apr · 2018]

Example 4ProbabilityMODERATE
Let AA and BB be two events such that P(AB)0.75P(A\cup B)\geq0.75 and 0.125P(AB)0.3750.125\leq P(A\cap B)\leq0.375.
What is the maximum value of P(A)+P(B)P(A)+P(B)?

[Q112 · Apr · 2024]

Example 5ProbabilityMODERATE
If P(AB)=56P(A\cup B)=\frac{5}{6}, P(AB)=13P(A\cap B)=\frac{1}{3} and P(not A)=12P(\text{not }A)=\frac{1}{2}, then which one of the following is not\textbf{\text{not}} correct?

[Q103 · Apr · 2020]

Drill every past-year question on this subtopic

11 questions from the bank — paginated, with cart and Word-export support.

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