Mathematics Class 12 — Part…
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Probability Exercises and Summary

🎓 Class 12 Mathematics CBSE Theory Ch 13 — Probability ⏱ ~15 min
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This MCQ module is based on: Probability Exercises and Summary

This mathematics assessment will be based on: Probability Exercises and Summary
Targeting Class 12 level in Calculus, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 13.1 — Conditional probability
1. P(E)=0.6, P(F)=0.3, P(E∩F)=0.2. Find P(E|F) and P(F|E).
P(E|F)=0.2/0.3=2/3. P(F|E)=0.2/0.6=1/3.
3. P(A)=8/13, P(B)=5/13, P(A∪B)=11/13. Find P(A∩B), P(A|B), P(B|A).
P(A∩B)=8/13+5/13−11/13=2/13. P(A|B)=(2/13)/(5/13)=2/5. P(B|A)=(2/13)/(8/13)=1/4.
5. A coin is tossed three times. \(E\) = "tail on second toss", \(F\) = "tail on first and head on third". Find P(E|F).
F = {THH, TTH}. E∩F = {TTH}. P(F)=2/8=1/4. P(E∩F)=1/8. P(E|F)=(1/8)/(1/4)=1/2.
10. A black die and a red die are rolled. Given the sum is 8, find P(red = 4).
F = "sum=8" = {(2,6),(3,5),(4,4),(5,3),(6,2)}: 5 outcomes. E∩F = "red=4 and sum=8" → black=4 → {(4,4)}: 1 outcome. P(E|F) = 1/5.
Exercise 13.2 — Multiplication theorem and independence
1. A and B are independent: P(A)=3/5, P(B)=1/5. Find P(A∩B).
P(A∩B) = (3/5)(1/5) = 3/25.
2. Two cards drawn without replacement from a deck. Find P(both red).
P(R₁) = 26/52 = 1/2. P(R₂|R₁) = 25/51. P(both) = (1/2)(25/51) = 25/102.
5. A die is rolled. \(E\) = "shows even", \(F\) = "shows ≥ 4". Test independence.
P(E)=3/6=1/2; P(F)=3/6=1/2; E∩F={4,6}, P=2/6=1/3. P(E)P(F)=1/4 ≠ 1/3. NOT independent.
9. P(A) = 0.3, P(B) = 0.6. A, B independent. Find P(A∩B), P(A∪B), P(A|B), P(B|A).
P(A∩B)=0.18. P(A∪B)=0.3+0.6−0.18=0.72. P(A|B)=0.18/0.6=0.3 (=P(A) ✓ independence). P(B|A)=0.18/0.3=0.6 (=P(B) ✓).
12. A die is tossed thrice. Find probability of (a) all even, (b) sum is even.
(a) P(even on each toss)=1/2; independent: (1/2)³=1/8. (b) Sum even ⇔ even count of odds. By symmetry/binomial: 4/8 = 1/2.
Exercise 13.3 — Total Probability and Bayes' Theorem
1. An urn has 5 red and 5 black balls. A red ball is drawn and 2 additional balls of the same colour are added. Then a second ball is drawn. Find P(2nd is red).
After drawing red and adding 2 reds: 6 red + 5 black = 11 balls; P(R₂|R₁) = 6/11. After drawing black and adding 2 blacks: 5 red + 6 black; P(R₂|B₁) = 5/11. Total: P(R₂) = (5/10)(6/11) + (5/10)(5/11) = (30+25)/110 = 55/110 = 1/2.
3. Of the students in a college, 60% are girls and 40% boys. 50% of girls and 30% of boys reside in hostel. A student is chosen at random and lives in hostel. Find probability the student is a girl.
P(G|H) = P(G)P(H|G) / [P(G)P(H|G) + P(B)P(H|B)] = 0.6·0.5 / (0.6·0.5 + 0.4·0.3) = 0.30/0.42 = 5/7.
4. P(A test correctly diagnoses HIV) = 99% sensitivity AND 99% specificity. Prevalence = 0.1%. Find P(HIV | test+).
P(HIV|+) = (0.001·0.99)/(0.001·0.99 + 0.999·0.01) = 0.00099/0.01098 ≈ 0.0902 ≈ 9%. Surprisingly low for a 99% test, due to base rate.
5. A bag has 4 white and 2 black; another has 3 white and 5 black. One bag is chosen at random and a ball drawn. Given the ball is white, find P(it came from the second bag).
P(B₂|W) = (1/2·3/8)/(1/2·4/6 + 1/2·3/8) = (3/16)/(2/6 + 3/16) = (3/16)/(16/48 + 9/48) = (3/16)·(48/25) = 9/25.
11. A bag contains 4 white and 6 black balls; another bag contains 7 white and 3 black. One ball is transferred from first to second; then a ball is drawn from second. Find P(white).
If transferred is W (P=4/10=2/5): bag 2 has 8W, 3B → P(W₂|W) = 8/11. If transferred is B (P=3/5): 7W, 4B → P(W₂|B) = 7/11. Total: (2/5)(8/11) + (3/5)(7/11) = 16/55 + 21/55 = 37/55.
Miscellaneous Exercise — selected
3. A bag contains tickets numbered 1 to 50. One is drawn. Find P(prime).
Primes ≤ 50: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 = 15 primes. P = 15/50 = 3/10.
9. A doctor visits a patient on each visit either: (i) walks (prob 0.3), (ii) cycles (0.4), (iii) drives (0.3). Probability of being late: 0.25, 0.30, 0.10 respectively. He arrives late. Find probability he came by car.
P(L) = 0.3·0.25 + 0.4·0.30 + 0.3·0.10 = 0.075 + 0.12 + 0.03 = 0.225. P(car|L) = 0.03/0.225 = 2/15.
Activity: Build a Bayesian Update Calculator
L4 Analyse
Materials: Paper, pen.
Predict: Given prior 60% chance of rain and a new piece of evidence with likelihood ratio LR = P(E|R)/P(E|R'), compute the posterior in one step.
  1. Setup: prior odds = P(R)/P(R'). E.g. 0.6/0.4 = 1.5.
  2. Likelihood ratio: LR = P(evidence | R) / P(evidence | R').
  3. Posterior odds = prior odds × LR (the "Bayes' rule in odds form").
  4. Posterior probability = (posterior odds) / (1 + posterior odds).
  5. Example: prior 0.6, LR = 4 (evidence is 4× more likely under R). Posterior odds = 1.5 × 4 = 6. Posterior probability = 6/7 ≈ 0.857.
The "odds form" of Bayes' rule is the cleanest way to chain multiple pieces of evidence: just multiply likelihood ratios. This is how naive Bayes spam classifiers and many real-world Bayesian systems actually work in practice.

Consolidation Competency-Based Questions

Scenario: A cookie jar has 3 chocolate-chip and 2 oatmeal cookies. Two are taken out one at a time without replacement.
Q1. P(both chocolate-chip):
L3 Apply
Answer: P(C₁) = 3/5; P(C₂|C₁) = 2/4 = 1/2. P(both) = (3/5)(1/2) = 3/10.
Q2. P(at least one oatmeal):
L3 Apply
Answer: Complement: 1 − P(both chocolate) = 1 − 3/10 = 7/10.
Q3. (T/F) "If P(A|B) > P(A), then 'B occurring' is positive evidence for A." Justify.
L5 Evaluate
True. P(A|B) > P(A) means knowing B raises confidence in A — positive evidence in the Bayesian sense. Equivalently P(B|A) > P(B). This is the formal definition of "B confirms A".
Q4. Apply: a coin is biased with P(H) = 0.6. Tossed 4 times. Find P(at least 1 head).
L4 Analyse
Solution: P(no heads) = (0.4)⁴ = 0.0256. P(at least 1) = 1 − 0.0256 = 0.9744.
Q5. Design: in a population, 1% have a rare condition. A test detects 95% true positives and 5% false positives. Construct a 10000-person table and read off P(condition | positive test).
L6 Create
Solution: 100 have it (95 test+); 9900 don't (495 test+). Total positives = 95+495=590. P(condition|+) = 95/590 ≈ 0.161 (≈16%). The "natural-frequencies" presentation makes Bayes intuitive without formulas.

Consolidation Assertion–Reason

Assertion (A): If E and F are independent, P(E∩F) = P(E)P(F).
Reason (R): Independence is defined precisely by the multiplicative formula P(E∩F) = P(E)P(F).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Tautological — R is the definition of A.
Assertion (A): A 99%-accurate test for a 1-in-10000 disease has high reliability for positives.
Reason (R): Bayes' theorem accounts for the base rate, which dominates when prevalence is small.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d). A is FALSE — for rare diseases the posterior is LOW despite a 99% test (this is the celebrated base-rate fallacy). R is true and explains why A is false.

Chapter Summary

Key concepts
  • Conditional probability: \(P(E|F)=P(E\cap F)/P(F)\), \(P(F)>0\).
  • Multiplication theorem: \(P(E\cap F)=P(E)P(F|E)=P(F)P(E|F)\). Generalises to n events.
  • Independence: \(P(E\cap F)=P(E)P(F)\). Equivalent: \(P(E|F)=P(E)\).
  • Mutually exclusive ≠ independent (with positive probabilities).
  • Partition: events \(E_1,\ldots,E_n\) pairwise disjoint and exhaustive.
  • Total probability: \(P(A)=\sum_i P(E_i)P(A|E_i)\).
  • Bayes' theorem: \(P(E_i|A)=\dfrac{P(E_i)P(A|E_i)}{\sum_j P(E_j)P(A|E_j)}\).
  • Prior → Posterior: Bayes converts prior beliefs into posterior via likelihood and evidence.

Historical Note

The mathematical theory of probability began with the famous 1654 correspondence between Blaise Pascal and Pierre de Fermat, sparked by gambler Chevalier de Méré's questions about dice. Christiaan Huygens (1657), Jacob Bernoulli (1713), and Abraham de Moivre (1718) developed the early theory.

Reverend Thomas Bayes (1701–1761), an English mathematician and Presbyterian minister, wrote his celebrated essay "An Essay towards solving a Problem in the Doctrine of Chances", published posthumously in 1763 by his friend Richard Price. The essay proved a special case of what we now call Bayes' theorem. Pierre-Simon Laplace (1812) independently rediscovered and dramatically generalised the result, applying it to celestial mechanics, population statistics, and the reliability of court verdicts.

The 20th century saw A.N. Kolmogorov's axiomatisation (1933), the rise of frequentist statistics (Fisher, Neyman), and the eventual triumphant return of Bayesian methods (Jeffreys, Jaynes, Lindley) — now powering machine learning, Bayesian networks, particle filters, MCMC simulation, and the AI revolution.

Frequently Asked Questions — Probability Exercises and Summary

What is the chapter summary of Class 12 Maths Probability?
Conditional probability, multiplication theorem, independence, total probability theorem, and Bayes' theorem — converting prior to posterior using evidence.
Who developed Bayes' theorem?
Reverend Thomas Bayes (1763 posthumous), independently rediscovered and generalised by Laplace (1812).
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