Introduction — quick memory-first summary
These advanced Probability notes cover everything you need: basic axioms, conditional probability, independence, Bayes' theorem, combinatorics, discrete distributions (Binomial & Poisson), expectation & variance, and problem-solving heuristics aimed to make the chapter unforgettable in one read.
Why this page?
- Memory-first layout: definition → theorem → proof → diagram → example → mnemonic.
- SVG diagrams (Venn, tree) for visual intuition; step-by-step solved examples for lifetime recall.
- SEO-ready: JSON-LD for Course/Breadcrumb/FAQ, unique meta (150–160 chars), H1/H2/H3 structure.
Core definitions & notation
- Experiment & sample space: sample space S is set of all outcomes; events are subsets of S.
- Probability axioms: For event A: 0 ≤ P(A) ≤ 1; P(S)=1; If A,B disjoint then P(A∪B)=P(A)+P(B).
- Conditional probability: P(A|B)=P(A∩B)/P(B) provided P(B)>0.
- Independence: A and B independent iff P(A∩B)=P(A)P(B).
- Bayes' theorem: P(A_i|B)=P(B|A_i)P(A_i)/Σ_j P(B|A_j)P(A_j) for partition {A_j}.
Theorems & proofs
Theorem — Addition rule (general)
For any events A and B: P(A∪B)=P(A)+P(B)-P(A∩B). Proof by partitioning sample space into disjoint parts.
Theorem — Multiplication rule & chain rule
P(A∩B)=P(A)P(B|A). For three events: P(A∩B∩C)=P(A)P(B|A)P(C|A∩B). Proof uses repeated conditional definitions.
SVG Diagrams (Venn, Tree & distribution sketch)
Formula & quick table
| Concept | Formula | When to use |
|---|---|---|
| Addition rule | P(A∪B)=P(A)+P(B)-P(A∩B) | Any two events |
| Conditional | P(A|B)=P(A∩B)/P(B) | Given B occurred |
| Independence | P(A∩B)=P(A)P(B) | Check if product rule holds |
| Bayes | P(A_i|B)=P(B|A_i)P(A_i)/Σ_j P(B|A_j)P(A_j) | Inverse probabilities |
| Binomial | P(X=k)=C(n,k)p^k(1-p)^{n-k} | n independent trials |
| Poisson (approx) | P(X=k)=e^{-λ}λ^k/k! | Rare events, large n |
| Expectation | E[X]=Σ x P(X=x) | Mean of discrete distribution |
| Variance | Var(X)=E[X^2]-(E[X])^2 | Dispersion |
परिचय — सारांश
ही Probability नोट्स axioms, conditional probability, Bayes, विभाजन, Binomial व Poisson वितरण, अपेक्षा व विचलन यांचा समावेश करतात — एक दगडात सर्व ज्ञान, सोप्या उदाहरणांसह.
8 Solved Examples — Step-by-step
Example 1 — Simple probability & complement
Problem: From a pack of 52 cards, one card is drawn. Find probability it is not an ace.
Solution: P(ace)=4/52=1/13. So P(not ace)=1 - 1/13 = 12/13.
Example 2 — Conditional probability (drawing without replacement)
Problem: Two cards drawn without replacement. P(second is ace | first is ace)?
Solution: Given first ace, remaining deck has 51 cards with 3 aces; probability = 3/51 = 1/17.
Example 3 — Bayes' theorem (medical test)
Problem: Disease prevalence 1%. Test sensitivity 95% (P(+|D)=0.95), specificity 90% (P(-|¬D)=0.90). If test is +, find P(D|+).
Solution:
- P(D)=0.01, P(¬D)=0.99. P(+|D)=0.95, P(+|¬D)=0.10.
- P(+)=0.95*0.01 + 0.10*0.99 = 0.0095 + 0.099 = 0.1085.
- P(D|+) = 0.0095 / 0.1085 ≈ 0.0876 ≈ 8.76%.
Example 4 — Binomial distribution
Problem: A fair coin tossed 5 times. P(get exactly 3 heads)?
Solution: n=5, k=3, p=0.5. P = C(5,3)(0.5)^3(0.5)^2 = 10 * (0.5)^5 = 10/32 = 5/16.
Example 5 — Poisson approx
Problem: Calls arrive at rate 2 per hour. Probability of exactly 3 calls in an hour?
Solution: λ=2. P(X=3)=e^{-2}2^3/3! = e^{-2}*8/6 ≈ (0.1353)*1.333 = 0.1804 (approx).
Example 6 — Independence check
Problem: Toss two fair coins. A = first is head, B = second is head. Are A and B independent?
Solution: P(A)=1/2, P(B)=1/2, P(A∩B)=1/4 = (1/2)*(1/2) ⇒ independent.
Example 7 — Conditional with tree
Problem: Urn1 has 2 red,3 black; Urn2 has 1 red,1 black. Choose urn1 with prob 0.6 else urn2. Draw one ball; find P(draw red).
Solution:
- P(red) = 0.6*(2/5) + 0.4*(1/2) = 0.6*0.4 + 0.4*0.5 = 0.24 + 0.2 = 0.44.
Example 8 — Expectation & variance
Problem: X takes values 0,1,2 with probabilities 0.2,0.5,0.3 respectively. Find E[X] and Var(X).
Solution:
- E[X]=0*0.2 +1*0.5 +2*0.3 = 0 +0.5 +0.6 =1.1.
- E[X^2]=0*0.2 +1*0.5 +4*0.3 = 0 +0.5 +1.2 =1.7. Var(X)=1.7 - (1.1)^2 =1.7 -1.21 =0.49.
Practice Questions
- Two dice rolled. P(sum is 8)?
- From 3 red & 4 blue balls, draw 2 without replacement. P(both red)?
- Given P(A)=0.3, P(B)=0.4, P(A∩B)=0.12. Find P(A∪B) and P(A|B).
- Binomial: n=10, p=0.2. P(at least one success)?
- Bayes: Multiple machines produce parts with different defect rates — compute posterior defect probability given observed defect.
Answer Key (brief)
- Ways to get sum 8: (2,6),(3,5),(4,4),(5,3),(6,2) → 5/36.
- P = C(3,2)/C(7,2) = 3 / 21 = 1/7.
- P(A∪B)=0.3+0.4-0.12=0.58. P(A|B)=0.12/0.4=0.3.
- P(at least one) = 1 - P(none) = 1 - (0.8)^{10} ≈ 1 - 0.1074 = 0.8926.
- Use Bayes: compute P(defect|machine) contributions and normalise — see worked steps on page.
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