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14.2 Axiomatic Approach to Probability

🎓 Class 11 Mathematics CBSE Theory Ch 14 — Probability ⏱ ~15 min
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This MCQ module is based on: 14.2 Axiomatic Approach to Probability

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Targeting Class 11 level in Probability, with Advanced difficulty.

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14.2 Axiomatic Approach to Probability

In Class 9 you used the empirical approach: \(P(A)\approx (\text{number of trials in which }A\text{ occurred})/(\text{total trials})\). For a coherent theory we need axioms that fix what "probability" means once and for all. Andrey Kolmogorov? gave such axioms in 1933, completing the modern theory.

Kolmogorov's Axioms
A probability on a sample space \(S\) is a function \(P\) assigning to each event \(A\subseteq S\) a real number \(P(A)\) satisfying:
  1. \(P(A)\ge 0\) for every event \(A\). (Non-negativity)
  2. \(P(S)=1\). (Normalisation)
  3. For mutually exclusive events \(A_1,A_2,\ldots\) (\(A_i\cap A_j=\varnothing\) for \(i\ne j\)), \[P(A_1\cup A_2\cup\cdots)=P(A_1)+P(A_2)+\cdots\quad\text{(Countable additivity)}.\]

Immediate consequences

Five basic theorems
  1. \(P(\varnothing)=0\) (impossible event).
  2. \(0\le P(A)\le 1\) for every event \(A\).
  3. \(P(A')=1-P(A)\) (complement law).
  4. If \(A\subseteq B\), then \(P(A)\le P(B)\) (monotonicity).
  5. Addition theorem: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). For mutually exclusive events the last term is 0.
Proof of the addition theorem
Decompose \(A\cup B\) into three disjoint pieces: \(A\setminus B,\ A\cap B,\ B\setminus A\). Apply axiom 3: \(P(A\cup B)=P(A\setminus B)+P(A\cap B)+P(B\setminus A)\).
Also \(P(A)=P(A\setminus B)+P(A\cap B)\) and \(P(B)=P(B\setminus A)+P(A\cap B)\). Add: \(P(A)+P(B)=P(A\setminus B)+P(B\setminus A)+2P(A\cap B)\).
Subtract \(P(A\cap B)\) to recover \(P(A\cup B)\). \(\square\)

14.2.1 Equally likely outcomes (the classical case)

Suppose the sample space \(S=\{\omega_1,\omega_2,\ldots,\omega_n\}\) is finite with \(n\) outcomes, all equally likely: \(P(\{\omega_i\})=p\) for every \(i\). Then by axioms 2 and 3, \(np=1\), so \(p=1/n\). Hence:

Classical probability formula
\[\boxed{\;P(A)=\dfrac{\text{number of outcomes in } A}{\text{total number of outcomes in } S}=\dfrac{|A|}{|S|}\;}\] This is the formula for "fair" experiments — uniform coins, balanced dice, well-shuffled decks.

14.2.2 Probability of "A or B"

For any two events \(A,B\):

\[P(A\cup B)=P(A)+P(B)-P(A\cap B).\]

For three events \(A,B,C\) (inclusion–exclusion):

\[P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(C\cap A)+P(A\cap B\cap C).\]

14.2.3 Probability of "not A"

Complement trick
\[P(\text{not }A)=P(A')=1-P(A).\] Often it is much easier to compute \(P(A')\) and subtract from 1. Example: "at least one head in 5 coin tosses" — its complement is "no heads at all", which has just 1 outcome out of 32.

Worked Examples

Example 7. A coin is tossed twice. Find the probability that at least one head appears.
\(S=\{HH,HT,TH,TT\}\), \(|S|=4\), all equally likely. Event "at least one head" = \(\{HH,HT,TH\}\), size 3. \(P=3/4\). Or by complement: \(P=1-P(TT)=1-1/4=3/4\).
Example 8. A die is thrown. What is the probability of getting (i) a number less than 5; (ii) a multiple of 3?
\(|S|=6\). (i) \(\{1,2,3,4\}\), \(P=4/6=2/3\). (ii) \(\{3,6\}\), \(P=2/6=1/3\).
Example 9. One card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is (i) a king; (ii) a face card; (iii) a red queen?
\(|S|=52\). (i) 4 kings: \(P=4/52=1/13\). (ii) 12 face cards (J, Q, K of each suit): \(P=12/52=3/13\). (iii) 2 red queens (♦Q, ♥Q): \(P=2/52=1/26\).
Example 10. Two students Anil and Ashima appear in an exam. P(Anil qualifies) = 0.05, P(Ashima qualifies) = 0.10, P(both qualify) = 0.02. Find P(at least one qualifies) and P(neither qualifies).
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.05+0.10-0.02=0.13\). \(P(\text{neither})=1-P(A\cup B)=1-0.13=0.87\).
Example 11. A bag contains 9 discs numbered 1 to 9, of which 4 are red, 3 are blue, 2 are yellow. A disc is drawn at random. Find the probability that it will be (i) red; (ii) yellow; (iii) blue or red.
(i) \(4/9\). (ii) \(2/9\). (iii) \(P(B)+P(R)=3/9+4/9=7/9\) (mutually exclusive colours).
Example 12. Three coins are tossed once. Find the probability of getting (i) all heads; (ii) at most 2 heads; (iii) exactly 2 heads.
\(|S|=8\). (i) \(\{HHH\}\): \(P=1/8\). (ii) "at most 2 H" = NOT "all heads" = \(1-1/8=7/8\). (iii) \(\{HHT,HTH,THH\}\): \(P=3/8\).
Example 13. Two dice are rolled. Find the probability that the sum is (i) 8; (ii) at most 5; (iii) more than 9.
\(|S|=36\). (i) Sum 8: \((2,6),(3,5),(4,4),(5,3),(6,2)\) — 5 outcomes, \(P=5/36\). (ii) Sums 2,3,4,5: counts 1+2+3+4=10, \(P=10/36=5/18\). (iii) Sums 10,11,12: counts 3+2+1=6, \(P=6/36=1/6\).
Example 14. Out of 100 students two sections of 40 and 60 are formed. If you and your friend are in the same section, find the probability that the section has 40 students.
Total ways to form sections (selecting 40 from 100) = \(\binom{100}{40}\). Equivalently, given you and friend together, condition on which section both are in. Direct: P(both in 40-section) = (40/100)(39/99); P(both in 60-section) = (60/100)(59/99). Conditional on "same section", P(in 40-section) = (40·39)/((40·39)+(60·59))=1560/(1560+3540)=1560/5100=26/85. Hence \(P=26/85\). [This requires conditional reasoning — see Class 12.]
Example 15. From a deck of 52 cards, four cards are drawn. Find the probability that all are diamonds.
\(P=\dfrac{\binom{13}{4}}{\binom{52}{4}}=\dfrac{715}{270725}=\dfrac{11}{4165}\approx 0.0026\).
Activity: Verify the Addition Theorem with a Card Deck
L3 Apply
Materials: Pen, paper.
Predict: When drawing one card from a deck, P(card is red OR card is king) = ?
  1. Define A = "card is red" (26 outcomes), B = "card is king" (4 outcomes).
  2. A ∩ B = "card is a red king" = {♦K, ♥K}, 2 outcomes.
  3. By the addition theorem: P(A ∪ B) = 26/52 + 4/52 − 2/52 = 28/52 = 7/13.
  4. Cross-check directly: A ∪ B = "any red card OR any king" = 26 + 2 (the two black kings) = 28 outcomes. P = 28/52 = 7/13. ✓
  5. Why does subtracting P(A∩B) work? Because outcomes in A∩B were counted once by P(A) and again by P(B) — we must subtract once to undo the double-count.
The addition theorem prevents double counting. With three events, you double-count three pairs, so you subtract them; but then triple intersections were over-corrected, so you add them back. This is the inclusion–exclusion principle, generalising to any number of sets.

Competency-Based Questions

Scenario: A weather model assigns: P(rain) = 0.6, P(wind) = 0.5, P(rain AND wind) = 0.3.
Q1. P(rain or wind):
L3 Apply
  • (a) 0.5
  • (b) 0.7
  • (c) 0.8
  • (d) 1.1
Answer: (c) 0.8. 0.6 + 0.5 − 0.3 = 0.8.
Q2. P(neither rain nor wind):
L3 Apply
Answer: P((R ∪ W)') = 1 − 0.8 = 0.2.
Q3. P(rain only, no wind):
L4 Analyse
Answer: P(R \\ W) = P(R) − P(R ∩ W) = 0.6 − 0.3 = 0.3.
Q4. (T/F) "If P(A) = 0.7 and P(B) = 0.4, then P(A ∩ B) ≥ 0.1." Justify.
L5 Evaluate
True. P(A ∪ B) ≤ 1, and P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1.1 − P(A ∩ B). So P(A ∩ B) ≥ 0.1. (Lower bound called the Bonferroni inequality.)
Q5. Design: from a deck of 52 cards, two are drawn without replacement. Find P(both are aces). Also P(at least one ace).
L6 Create
Solution: P(both aces) = \(\binom{4}{2}/\binom{52}{2}=6/1326=1/221\). For "at least one ace", complement = "no aces" = \(\binom{48}{2}/\binom{52}{2}=1128/1326=188/221\). So P(≥1 ace) = 1 − 188/221 = 33/221.

Assertion–Reason Questions

Assertion (A): \(P(A')=1-P(A)\).
Reason (R): \(A\) and \(A'\) are mutually exclusive and exhaustive, so \(P(A)+P(A')=P(S)=1\).
(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). R is precisely the proof of A.
Assertion (A): If \(A,B\) are mutually exclusive, \(P(A\cup B)=P(A)+P(B)\).
Reason (R): The general formula is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), and mutual exclusion makes \(P(A\cap B)=0\).
(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). R is the simplification that yields A.
Assertion (A): Probability of an impossible event is 0.
Reason (R): The empty set has no outcomes, so the count is 0.
(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). Both true; the empty set is the impossible event and gets probability 0.

Frequently Asked Questions

What are the three axioms of probability?
(1) For every event A, P(A) ≥ 0. (2) P(S) = 1. (3) For mutually exclusive events A and B, P(A ∪ B) = P(A) + P(B). These are Kolmogorov's axioms (1933).
What is the addition theorem of probability?
P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For mutually exclusive events, A ∩ B = ∅, so the formula simplifies to P(A ∪ B) = P(A) + P(B).
What is P(A')?
P(A') = 1 − P(A), because A and A' are mutually exclusive and exhaustive: P(A) + P(A') = P(S) = 1.
What is the formula for equally likely outcomes?
If a sample space has n equally likely outcomes and an event A consists of m of them, then P(A) = m/n.
What is the probability of getting at least one head in 3 coin tosses?
P(at least one head) = 1 − P(no heads) = 1 − P(TTT) = 1 − 1/8 = 7/8.
What is the probability of an impossible event?
P(∅) = 0. The empty event corresponds to no outcome at all and so it never occurs.
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