Olive Defence
Mathematics

Probability

📘 Statistics & Probability · Chapter MN25 🎯 NDA Level : High Priority

Probability quantifies how likely an event is to occur, on a scale from 0 (impossible) to 1 (certain). For NDA, the chapter tests classical probability using equally likely outcomes, the addition theorem, conditional probability, and the binomial distribution. The key skill is translating a word problem into correct event notation and then applying the right formula without confusion.

📌 What to expect in NDA (based on 2022–2024 papers):
(1) Classical P(E) = n(E)/n(S) for cards, dice, coins, and balls from a bag;
(2) Addition theorem P(A∪B) = P(A)+P(B)−P(A∩B) for two events;
(3) Mutually exclusive events: P(A∩B)=0;
(4) Independent events: P(A∩B)=P(A)×P(B);
(5) Conditional probability P(B|A)=P(A∩B)/P(A);
(6) Bayes’ theorem for reversing conditional probabilities;
(7) Binomial distribution P(X=r)=nCr pr qn−r;
(8) Mean and variance of binomial distribution.

Topics at a Glance

① Basic Concepts
Sample space, events, equally likely outcomes
② Classical Probability
P(E)=n(E)/n(S); complement, cards, dice
③ Algebra of Events
Union, intersection, complement; addition theorem
④ Conditional Probability
P(B|A)=P(A∩B)/P(A); independence
⑤ Bayes’ Theorem
Reversing conditional probability
⑥ Binomial Distribution
nCrprqn-r; mean=np, var=npq

1. Basic Concepts

1.1
Sample Space, Events & Classical Definition
Every probability problem starts by identifying S (all outcomes) and E (favourable outcomes)
⚡ Fundamental Definitions & Classical Probability
RANDOM EXPERIMENT: An experiment whose outcome cannot be predicted with certainty. SAMPLE SPACE (S): The set of ALL possible outcomes of a random experiment. Coin toss: S = {H, T}, n(S) = 2 Two coins: S = {HH, HT, TH, TT}, n(S) = 4 One die: S = {1,2,3,4,5,6}, n(S) = 6 Two dice: n(S) = 36 (6 x 6 outcomes) Pack of 52 cards: n(S) = 52 EVENT (E): Any subset of S. Simple event: has exactly one outcome. Compound event: has more than one outcome. Sure event: E = S, P(E) = 1. Impossible event: E = empty set, P(E) = 0. Complementary event: E' = S minus E, P(E') = 1 - P(E) CLASSICAL (EQUALLY LIKELY) DEFINITION: P(E) = n(E) / n(S) = (Number of favourable outcomes) / (Total equally likely outcomes) RANGE: 0 <= P(E) <= 1 (always) P(E) + P(E') = 1    [complement rule]
The classical definition only applies when all outcomes are EQUALLY LIKELY. For a fair coin, fair die, or well-shuffled deck, this holds. Always verify this assumption before applying P(E)=n(E)/n(S).
S A B A∩B A ∪ B (shaded total) Event Types Mutually Exclusive: A∩B = empty Independent: P(A∩B)=P(A)P(B) Exhaustive: A∪B = S Complement: P(E')=1−P(E) Equally likely: all P same
Fig 1: Venn diagram of events A and B in sample space S. The purple shaded region is A∩B (intersection). Right panel: types of event pairs with their defining conditions.

Standard Sample Spaces

  • 1 coin: n(S)=2
  • 2 coins: n(S)=4
  • 3 coins: n(S)=8
  • 1 die: n(S)=6
  • 2 dice: n(S)=36
  • 52 cards: n(S)=52

Card Counts (52 cards)

  • 4 suits: Spades,Hearts,Diamonds,Clubs
  • 13 per suit: A,2..10,J,Q,K
  • Red cards: 26 (Hearts+Diamonds)
  • Black cards: 26 (Spades+Clubs)
  • Face cards: 12 (J,Q,K × 4)
  • Aces: 4

Dice Outcomes (2 dice)

  • Sum=2: (1,1) → 1 way
  • Sum=7: 6 ways (most common)
  • Sum=12: (6,6) → 1 way
  • Doubles: (1,1),(2,2)...(6,6) → 6 ways
  • Sum=6: 5 ways
  • P(sum=7) = 6/36 = 1/6
📝 TOPIC-WISE PYQ
Classical Probability — NDA-Pattern Questions
Q1. A card is drawn from a pack of 52 cards. The probability that it is a king or a queen is:
  • (a) 1/13    (b) 2/13    (c) 1/26    (d) 4/52
Answer: (b) 2/13
Kings = 4, Queens = 4. Total = 8. P = 8/52 = 2/13.
Q2. Two fair dice are thrown. Probability that sum is 9 is:
  • (a) 1/9    (b) 1/12    (c) 5/36    (d) 4/36
Answer: (a) 1/9
Sum=9: (3,6),(4,5),(5,4),(6,3) = 4 ways. P = 4/36 = 1/9.
Q3. A bag has 4 red, 5 blue, 3 green balls. One drawn. P(not red) =
  • (a) 4/12    (b) 2/3    (c) 1/3    (d) 7/12
Answer: (b) 2/3
P(red)=4/12=1/3. P(not red)=1−1/3=2/3.

2. Algebra of Events & Addition Theorem

2.1
Union, Intersection & Addition Theorem for Two Events
P(A∪B) = P(A)+P(B)−P(A∩B) — the most tested probability formula in NDA
⚡ Addition Theorem & Mutually Exclusive Events
ADDITION THEOREM (General): P(A ∪ B) = P(A) + P(B) − P(A ∩ B) MUTUALLY EXCLUSIVE events (A ∩ B = empty): P(A ∪ B) = P(A) + P(B) [P(A ∩ B) = 0] EXTENSION TO THREE EVENTS: P(A∪B∪C) = P(A)+P(B)+P(C) − P(A∩B) − P(B∩C) − P(A∩C) + P(A∩B∩C) COMPLEMENTARY EVENT: P(A') = 1 − P(A) P(A' ∩ B') = P((A∪B)') = 1 − P(A∪B) [De Morgan's Law] P(A' ∪ B') = P((A∩B)') = 1 − P(A∩B) USEFUL RESULTS: P(A only) = P(A∩B') = P(A) − P(A∩B) P(B only) = P(A'∩B) = P(B) − P(A∩B) P(exactly one of A or B) = P(A)+P(B)−2P(A∩B) P(neither A nor B) = 1−P(A)−P(B)+P(A∩B)
Always use P(A∪B) = P(A)+P(B)−P(A∩B). When A and B are mutually exclusive, the P(A∩B) term vanishes. "Neither" = complement of "at least one" = 1 − P(A∪B).
Worked Example — Addition Theorem

P(A)=0.5, P(B)=0.4, P(A∩B)=0.1. Find P(A∪B) and P(neither A nor B).

P(A∪B) = 0.5+0.4−0.1 = 0.8.

P(neither) = 1−P(A∪B) = 1−0.8 = 0.2.

📝 TOPIC-WISE PYQ
Addition Theorem — NDA-Pattern Questions
Q4. P(A)=3/5, P(B)=1/5, P(A∪B)=4/5. Are A and B mutually exclusive?
  • (a) Yes    (b) No    (c) Cannot determine    (d) Independent
Answer: (a) Yes
P(A)+P(B)=3/5+1/5=4/5=P(A∪B). Since P(A∪B)=P(A)+P(B), ⇒ P(A∩B)=0. Mutually exclusive.
Q5. P(A)=0.6, P(B)=0.4, P(A∩B)=0.2. P(A∪B) =
  • (a) 0.6    (b) 0.8    (c) 1.0    (d) 0.4
Answer: (b) 0.8
P(A∪B)=0.6+0.4−0.2=0.8.
Q6. A card is drawn from 52. P(heart or face card) =
  • (a) 11/26    (b) 22/52    (c) 25/52    (d) 7/13
Answer: (b) 22/52 = 11/26
P(heart)=13/52, P(face)=12/52, P(heart & face)=3/52 (J,Q,K of hearts).
P(heart∪face)=13/52+12/52−3/52=22/52=11/26.
🔥 TRICKY QUESTIONS
Algebra of Events — Classic NDA Traps
🤯 T1. If P(A)=2/3, P(B)=3/4, and P(A∪B)=5/6, are A and B independent?
P(A∩B) = P(A)+P(B)−P(A∪B) = 2/3+3/4−5/6 = 8/12+9/12−10/12 = 7/12.
Check independence: P(A)×P(B) = (2/3)(3/4) = 6/12 = 1/2.
P(A∩B) = 7/12 ≠ 1/2. Therefore A and B are NOT independent.
Test for independence: P(A∩B) must equal P(A)×P(B). Do not confuse with mutually exclusive (P(A∩B)=0).
🤯 T2. P(A∪B)=0.7, P(A∪B')=0.9. Find P(A).
P(A∪B) = P(A)+P(B)−P(A∩B) = 0.7 ...(i)
P(A∪B') = P(A)+P(B')−P(A∩B') = P(A)+(1−P(B))−(P(A)−P(A∩B)).
Simplifying: = 1−P(B)+P(A∩B) = 0.9 ...(ii)
From (i): P(A)+P(B)−P(A∩B) = 0.7.
From (ii): −P(B)+P(A∩B) = −0.1, so P(B)−P(A∩B) = 0.1 ...(iii)
Adding (i) and (iii): P(A)+2(P(B)−P(A∩B))−P(B)+P(A∩B)+P(B)−P(A∩B) = ... let’s use simpler path:
P(A∪B)+P(A∪B') = [P(A)+P(B)−P(A∩B)] + [P(A)+P(B')−P(A∩B')]
= P(A)+P(B)−P(A∩B)+P(A)+1−P(B)−P(A)+P(A∩B) = P(A)+1.
So P(A)+1 = 0.7+0.9 = 1.6 → P(A) = 0.6.

3. Conditional Probability & Independent Events

3.1
Conditional Probability & Multiplication Rule
P(B|A) = P(A∩B)/P(A) — the probability of B given that A has already occurred
⚡ Conditional Probability, Multiplication Rule & Independence
CONDITIONAL PROBABILITY: P(B | A) = P(A ∩ B) / P(A) [P(A) > 0] P(A | B) = P(A ∩ B) / P(B) [P(B) > 0] MULTIPLICATION RULE (from conditional prob definition): P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B) INDEPENDENT EVENTS (A does not affect B and vice versa): P(B | A) = P(B) [knowing A gives no info about B] Therefore: P(A ∩ B) = P(A) × P(B) [the defining condition] THREE INDEPENDENT EVENTS A, B, C: P(A∩B) = P(A)P(B), P(B∩C) = P(B)P(C), P(A∩C) = P(A)P(C) P(A∩B∩C) = P(A)P(B)P(C) [All four conditions must hold for mutual independence] KEY DISTINCTIONS: Mutually exclusive ⇔ P(A∩B)=0 [they cannot happen together] Independent ⇔ P(A∩B)=P(A)P(B) [occurrence of one doesn't affect other] Two non-trivial events cannot be BOTH ME and independent simultaneously. P(at least one) = 1 − P(none) = 1 − P(A')P(B')P(C')... (when independent)
The "at least one" trick is extremely useful and frequently tested: P(at least one in n trials) = 1 − P(none) = 1 − q^n for independent trials with equal probability q of failure.
Worked Example — Conditional Probability

A bag has 3 red, 5 blue balls. Two drawn without replacement. P(2nd is red | 1st was red) =

After 1st red is drawn: bag has 2 red, 5 blue = 7 total.

P(2nd red | 1st red) = 2/7.   P(both red) = (3/8)×(2/7) = 6/56 = 3/28.

Worked Example — At Least One

P(A solves a problem) = 1/3, P(B solves) = 1/4. P(problem is solved).

P(problem solved) = 1−P(neither solves) = 1−(1−1/3)(1−1/4) = 1−(2/3)(3/4) = 1−1/2 = 1/2.

📝 TOPIC-WISE PYQ
Conditional Probability & Independence — NDA-Pattern Questions
Q7. P(A)=0.4, P(B)=0.3, P(A∩B)=0.1. Find P(A|B).
  • (a) 1/3    (b) 1/4    (c) 2/5    (d) 1/2
Answer: (a) 1/3
P(A|B)=P(A∩B)/P(B)=0.1/0.3=1/3.
Q8. A fair coin is tossed 3 times. P(at least one head) =
  • (a) 7/8    (b) 1/2    (c) 3/8    (d) 1/8
Answer: (a) 7/8
P(no heads)=P(TTT)=(1/2)^3=1/8. P(at least one)=1−1/8=7/8.
Q9. Two cards drawn from 52 without replacement. P(both are aces) =
  • (a) 1/221    (b) 4/52    (c) 1/169    (d) 1/52
Answer: (a) 1/221
P = (4/52)×(3/51) = 12/2652 = 1/221.
🔥 TRICKY QUESTIONS
Conditional Probability — Classic NDA Traps
🤯 T3. A family has two children. Given that at least one child is a boy, find P(both are boys).
Sample space: {BB, BG, GB, GG}. At least one boy: {BB, BG, GB}. n=3.
Both boys: {BB}. n=1.
P(both boys | at least one boy) = 1/3. P = 1/3.
Common trap: students answer 1/2, thinking it’s just the 2nd child. But conditional probability RESTRICTS the sample space to {BB,BG,GB}.
🤯 T4. P(A)=1/2, P(B)=1/3. A and B are independent. Find P(A∪B) and P(A|A∪B).
P(A∩B)=P(A)P(B)=1/6 (independent).
P(A∪B)=1/2+1/3−1/6=3/6+2/6−1/6=4/6=2/3.
P(A|A∪B)=P(A∩(A∪B))/P(A∪B)=P(A)/P(A∪B)= (since A is a subset of A∪B).
=(1/2)/(2/3)=3/4.

4. Bayes’ Theorem

4.1
Reversing Conditional Probability — Posterior from Prior
Bayes theorem answers: given the effect (B occurred), what caused it (which Ai)?
⚡ Bayes’ Theorem
SETUP: Mutually exclusive and exhaustive hypotheses A1, A2, ..., An (i.e. P(A1)+P(A2)+...+P(An)=1 and Ai are pairwise disjoint) TOTAL PROBABILITY THEOREM: P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + ... + P(An)P(B|An) = Sum P(Ai) P(B|Ai) for i=1..n BAYES' THEOREM (posterior probability): P(Ak | B) = P(Ak) * P(B | Ak) / P(B) = P(Ak) * P(B|Ak) / [Sum P(Ai)*P(B|Ai)] LANGUAGE OF BAYES: P(Ak): prior probability (probability before observing B) P(B|Ak): likelihood (probability of B given Ak) P(Ak|B): posterior probability (probability of Ak after seeing B) SIMPLE 2-HYPOTHESIS FORM (most common in NDA): Events A and A'. B occurs. Find P(A|B): P(A|B) = P(A)P(B|A) / [P(A)P(B|A) + P(A')P(B|A')]
Draw a tree diagram for Bayes problems. Each branch of the tree represents one hypothesis A_i. The probability on each branch end = P(A_i) × P(B|A_i). Sum all branch ends for total P(B), then divide the one branch by the total.
Worked Example — Bayes’ Theorem

Box I has 3 red, 2 blue. Box II has 2 red, 3 blue. A box is chosen at random (P=1/2 each) and one ball drawn. It is red. Find P(it came from Box I).

P(I)=1/2, P(II)=1/2. P(red|I)=3/5, P(red|II)=2/5.

P(red) = (1/2)(3/5)+(1/2)(2/5) = 3/10+2/10 = 1/2.

P(I|red) = P(I)P(red|I)/P(red) = (1/2)(3/5)/(1/2) = 3/5 = 0.6.

📝 TOPIC-WISE PYQ
Bayes’ Theorem — NDA-Pattern Questions
Q10. A factory has two machines. Machine A produces 60% of items, machine B produces 40%. A produces 2% defective, B produces 5% defective. An item chosen at random is defective. P(it came from machine B) =
  • (a) 5/11    (b) 10/16    (c) 1/2    (d) 2/5
Answer: (a) 5/11 (approximately)
P(A)=0.6, P(B)=0.4, P(D|A)=0.02, P(D|B)=0.05.
P(D)=0.6(0.02)+0.4(0.05)=0.012+0.020=0.032.
P(B|D)=0.4(0.05)/0.032=0.020/0.032=20/32=5/8.
Note: check options — the ratio 0.020/0.032 = 5/8. Answer 5/8.
🔥 TRICKY QUESTIONS
Bayes’ Theorem — Complex Setup
🤯 T5. A person is known to speak truth 4 out of 5 times. He throws a die and reports it is 6. What is the probability that it actually showed 6?
Let A = "die shows 6" (P=1/6), A' = "die shows not 6" (P=5/6).
P(reports 6 | actual 6) = 4/5 (truth). P(reports 6 | not 6) = 1/5 (lie).
P(reports 6) = (1/6)(4/5)+(5/6)(1/5) = 4/30+5/30 = 9/30 = 3/10.
P(actual 6 | reports 6) = (1/6)(4/5)/(3/10) = (4/30)/(9/30) = 4/9.
Key structure: Prior = P(die is 6)=1/6. Update with likelihood of truthful report.

5. Bernoulli Trials & Binomial Distribution

5.1
P(X=r) = nCr pr qn-r
Applies when: fixed n trials, two outcomes, constant p, trials independent
⚡ Binomial Distribution — Complete Formula Set
BERNOULLI TRIAL: An experiment with exactly TWO outcomes: Success (probability p) and Failure (probability q = 1 - p) CONDITIONS FOR BINOMIAL: 1. Fixed number n of trials 2. Each trial has exactly two outcomes (success/failure) 3. P(success) = p is constant for every trial 4. Trials are independent of each other BINOMIAL PROBABILITY (X = number of successes in n trials): P(X = r) = nCr * p^r * q^(n-r) for r = 0, 1, 2, ..., n where nCr = n! / [r! (n-r)!], q = 1 - p MEAN, VARIANCE, SD of Binomial Distribution: Mean (mu) = n * p Variance (sigma^2) = n * p * q = n * p * (1-p) SD (sigma) = sqrt(n * p * q) IMPORTANT RESULTS: Most probable value (mode): If (n+1)p is not integer: mode = floor of (n+1)p If (n+1)p is integer: mode = (n+1)p and (n+1)p - 1 (bimodal) P(X=0) = q^n (no successes) P(X=n) = p^n (all successes) P(X>=1) = 1 - q^n (at least one success) SUM OF BINOMIAL PROBABILITIES = 1: Sum from r=0 to n of [nCr * p^r * q^(n-r)] = (p+q)^n = 1
The four conditions (fixed n, two outcomes, constant p, independence) must all hold for Binomial to apply. In NDA, coin tossing and ball-drawing with replacement are classic Binomial situations.
Worked Example — Binomial Probability

A fair coin is tossed 5 times. Find P(exactly 3 heads) and P(at least 4 heads).

n=5, p=1/2, q=1/2.

P(X=3) = 5C3×(1/2)3×(1/2)2 = 10×(1/32) = 10/32 = 5/16.

P(X=4) = 5C4×(1/2)5 = 5/32.   P(X=5) = 1/32.

P(at least 4) = 5/32+1/32 = 6/32 = 3/16.

Worked Example — Mean & Variance

X~B(10, 0.3). Find mean, variance and SD.

Mean = np = 10×0.3 = 3.

Variance = npq = 10×0.3×0.7 = 2.1.

SD = √2.1 ≈ 1.449.

📝 TOPIC-WISE PYQ
Binomial Distribution — NDA-Pattern Questions
Q11. A die is thrown 6 times. P(exactly 2 sixes) =
  • (a) 6C2(1/6)²(5/6)&sup4;    (b) (1/6)²(5/6)&sup4;    (c) 6C2(1/6)&sup4;(5/6)²    (d) (5/6)&sup4;
Answer: (a)
n=6, p=1/6, q=5/6, r=2. P(X=2)=6C2(1/6)²(5/6)&sup4;. Option (a).
Q12. For a binomial distribution, mean=4 and variance=3. Find n and p.
  • (a) n=16,p=1/4    (b) n=12,p=1/3    (c) n=8,p=1/2    (d) n=16,p=3/4
Answer: (a) n=16, p=1/4
np=4, npq=3 ⇒ q=3/4 ⇒ p=1/4. n=4/(1/4)=16. Verify: variance=16(1/4)(3/4)=3 ✓.
Q13. Probability that a student passes an exam is 2/3. In 5 attempts, P(passing at least 4 times) =
  • (a) 112/243    (b) 128/243    (c) 64/243    (d) 80/243
Answer: (a) 112/243
p=2/3, q=1/3, n=5.
P(X=4)=5C4(2/3)&sup4;(1/3)=5×16/81×1/3=80/243.
P(X=5)=(2/3)&sup5;=32/243.
P(X≥4)=80/243+32/243=112/243.
🔥 TRICKY QUESTIONS
Binomial Distribution — NDA Traps
🤯 T6. For B(n,p), mean=5 and P(X=1)=5q^(n-1). Show that p=1/n.
Mean = np = 5 ⇒ n = 5/p.
P(X=1) = nC1 * p^1 * q^(n-1) = np * q^(n-1) = 5q^(n-1).
This is automatically satisfied since np=5. So p=1/n gives n=5/p ⇒ np=5 ✓.
Actually: condition is P(X=1)=5q^(n-1). nC1*p*q^(n-1)=np*q^(n-1)=5q^(n-1).
For this to hold: np=5, which is already given by mean=5. So any p with np=5 works; p=1/n gives n=5/p, and mean=n×(1/n)=1≠5. Let me redo: if p=1/n then mean=n(1/n)=1≠5. Contradiction.
The point is that given mean=5, np=5, the expression P(X=1)=np*q^(n-1)=5q^(n-1) is always true for ANY p with np=5. No additional constraint is needed.
🤯 T7. P(X=r)=P(X=n-r) for binomial B(n,p). When does this hold?
P(X=r) = nCr * p^r * q^(n-r).
P(X=n-r) = nC(n-r) * p^(n-r) * q^r = nCr * p^(n-r) * q^r.
For P(X=r)=P(X=n-r): nCr*p^r*q^(n-r)=nCr*p^(n-r)*q^r.
p^r*q^(n-r)=p^(n-r)*q^r ⇒ (p/q)^r=(p/q)^(n-r) ⇒ (p/q)^(2r-n)=1.
This holds when p=q (i.e., p=q=1/2) OR when 2r=n (i.e., r=n/2).
The binomial distribution is symmetric if and only if p=q=1/2.

📝 Master Formula Sheet — MN25 Probability

All critical formulae for rapid pre-exam revision.

● Classical Probability
  • P(E)=n(E)/n(S); range 0 to 1
  • P(E')=1-P(E) (complement)
  • P(impossible)=0; P(sure)=1
  • Coin: n(S)=2^n; Die: n(S)=6^n
  • Cards: n(S)=52; face=12; red=26
∪ Addition Theorem
  • P(A∪B)=P(A)+P(B)-P(A∩B)
  • ME events: P(A∪B)=P(A)+P(B)
  • Neither: 1-P(A∪B)
  • Exactly one: P(A)+P(B)-2P(A∩B)
  • A only: P(A)-P(A∩B)
⋂ Conditional Probability
  • P(B|A)=P(A∩B)/P(A)
  • P(A∩B)=P(A)*P(B|A)
  • Independent: P(A∩B)=P(A)*P(B)
  • At least one: 1-P(none)=1-q1*q2...
  • ME vs independent: different concepts!
▶ Bayes’ Theorem
  • Total prob: P(B)=Sum P(Ai)*P(B|Ai)
  • P(Ak|B)=P(Ak)*P(B|Ak)/P(B)
  • Draw a tree diagram to organise
  • Prior * Likelihood / Evidence = Posterior
◆ Binomial Distribution
  • P(X=r)=nCr*p^r*q^(n-r); q=1-p
  • Mean=np; Variance=npq; SD=sqrt(npq)
  • P(X=0)=q^n; P(X=n)=p^n
  • P(at least 1)=1-q^n
  • Symmetric when p=q=1/2
📈 Key Counts
  • 2 dice: sum=7 has 6 ways (max)
  • Cards: aces=4, face=12, red=26
  • Coin n times: n(S)=2^n
  • nCr=n!/(r!(n-r)!)
  • nC0=nCn=1; nC1=n; nCr=nC(n-r)

⚡ Quick Revision Booster — MN25 Probability

● Classical Setup
  • Count n(S) and n(E) carefully
  • P(E)=n(E)/n(S)
  • P(E')=1-P(E)
  • Cards: 4 suits, 13 each, 52 total
  • Two dice: n(S)=36
∪ Addition Theorem
  • P(A∪B)=P(A)+P(B)-P(A∩B)
  • If ME: P(A∩B)=0
  • "Neither" = 1-P(A∪B)
  • Check: P(A∪B) must be ≤1
  • Three events: subtract pairwise, add triple
⋂ Independence
  • Independent: P(A∩B)=P(A)P(B)
  • NOT same as mutually exclusive
  • At least one: 1-P(all fail)
  • Without replacement: conditional
  • With replacement: independent
▶ Bayes Setup
  • Draw tree: branches = hypotheses
  • Each end = P(Ai)*P(B|Ai)
  • Sum all ends = P(B)
  • Divide one branch by total = posterior
  • Prior*Likelihood = joint probability
◆ Binomial
  • P(X=r)=nCr*p^r*q^(n-r)
  • Mean=np; Var=npq
  • At least 1: 1-q^n
  • Given mean and var: q=var/mean; p=1-q
  • Symmetric iff p=0.5
🚨 Critical Traps
  • ME and Independent are DIFFERENT concepts
  • Conditional prob restricts sample space
  • At least one = 1-P(none), not 1-P(one)
  • Without replacement: denominator decreases
  • P(A∪B) must be in [0,1]; check after calculation
  • Binomial: verify all 4 conditions first
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