Mathematics Class 12 — Part…
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Conditional Probability, Multiplication Theorem and Independent Events

🎓 Class 12 Mathematics CBSE Theory Ch 13 — Probability ⏱ ~15 min
🌐 Language: [gtranslate]

This MCQ module is based on: Conditional Probability, Multiplication Theorem and Independent Events

This mathematics assessment will be based on: Conditional Probability, Multiplication Theorem and Independent Events
Targeting Class 12 level in Calculus, with Advanced difficulty.

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13.1 Introduction

In Class 11 you learned probability through equally likely outcomes and the addition rule. This chapter extends the theory in two key directions: conditional probability? (probability under partial information) and independence? (when one event tells you nothing about another). The chapter culminates in Bayes' theorem — a single formula that powers everything from spam filters to medical diagnostics to modern AI.

Pierre de Fermat

1601 – 1665

French jurist and mathematician. In a famous 1654 correspondence with Blaise Pascal — sparked by gambler Chevalier de Méré's dice problems — Fermat and Pascal laid the foundations of probability theory. Fermat is also celebrated for his contributions to number theory, optics, and analytic geometry, and for the legendary "Fermat's Last Theorem" (proved by Andrew Wiles in 1995).

13.2 Conditional Probability

Conditional probability
For events \(E\) and \(F\) of a sample space \(S\) with \(P(F)\ne 0\), the conditional probability of \(E\) given \(F\) is \[\boxed{\;P(E|F)=\dfrac{P(E\cap F)}{P(F)}\;}\] Read: "probability of E given F". It is the probability of E in the reduced sample space where we already know F occurred.

Example: a fair die is rolled. Let \(E=\{\)an even number\(\}=\{2,4,6\}\), \(F=\{\)number ≥ 3\(\}=\{3,4,5,6\}\). Then \(P(E)=3/6=1/2\), \(P(F)=4/6=2/3\), \(E\cap F=\{4,6\}\), \(P(E\cap F)=2/6=1/3\). Conditional: \(P(E|F)=(1/3)/(2/3)=1/2\). The information "F occurred" did not change the probability of E here — these events happen to be independent.

13.2.1 Properties of conditional probability

Three key properties
For any events \(E, F, G\) (with the conditioning event having non-zero probability):
  1. \(P(S|F)=P(F|F)=1\); \(P(\varnothing|F)=0\).
  2. \(P(E'|F)=1-P(E|F)\) (complement law conditioned on F).
  3. \(P((E\cup G)|F)=P(E|F)+P(G|F)-P(E\cap G|F)\) (addition rule conditioned on F).

13.3 Multiplication Theorem on Probability

Multiplication theorem (two events)
Rearranging the conditional probability formula: \[\boxed{\;P(E\cap F)=P(F)\cdot P(E|F)=P(E)\cdot P(F|E)\;}\] both forms valid (with the corresponding probability non-zero). Generalises to \(n\) events: \[P(E_1\cap E_2\cap\cdots\cap E_n)=P(E_1)\cdot P(E_2|E_1)\cdot P(E_3|E_1\cap E_2)\cdots P(E_n|E_1\cap\cdots\cap E_{n-1}).\]

13.4 Independent Events

Independence (two events)
Events \(E\) and \(F\) are independent if the occurrence of one does not affect the probability of the other: \[P(E|F)=P(E)\quad\Longleftrightarrow\quad P(F|E)=P(F)\quad\Longleftrightarrow\quad \boxed{\;P(E\cap F)=P(E)\cdot P(F)\;}\] The third form is the standard test (handles even when one of P(E), P(F) is 0).
Mutually Exclusive ≠ Independent
These are very different concepts:
• Mutually exclusive: \(E\cap F=\varnothing\Rightarrow P(E\cap F)=0\) (cannot occur together).
• Independent: \(P(E\cap F)=P(E)\cdot P(F)\) (do not influence each other).
For events with non-zero probabilities, mutual exclusion implies one cancels out the other when given — they are maximally dependent, not independent.
Independence of derived events
If E and F are independent:
  • E and F' are independent.
  • E' and F are independent.
  • E' and F' are independent.
Independence is preserved by complementation.

Worked Examples

Example 1. A fair die is rolled. Find P(\(E\)|\(F\)) where \(E\) = "appearance of an odd number", \(F\) = "appearance of a number ≥ 4".
\(E=\{1,3,5\}, F=\{4,5,6\}\); \(E\cap F=\{5\}\). \(P(F)=3/6=1/2,\ P(E\cap F)=1/6\). \(P(E|F)=(1/6)/(1/2)=1/3\).
Example 2. A fair coin is tossed three times. Find \(P(E|F)\) where E = "head on the third toss", F = "heads on first two tosses".
\(F=\{HHH, HHT\}\), \(P(F)=2/8=1/4\). \(E\cap F=\{HHH\}\), \(P(E\cap F)=1/8\). \(P(E|F)=(1/8)/(1/4)=1/2\). (Coin tosses are independent — the first two don't affect the third.)
Example 3. \(P(A)=7/13,\ P(B)=9/13,\ P(A\cap B)=4/13\). Find \(P(A|B)\).
\(P(A|B)=P(A\cap B)/P(B)=(4/13)/(9/13)=4/9\).
Example 4. An urn contains 10 black and 5 white balls. Two balls are drawn without replacement. Find the probability that both are black.
Let A = "first black", B = "second black". P(A) = 10/15 = 2/3. After drawing one black, 9 black + 5 white = 14 left. P(B|A) = 9/14. By multiplication theorem: \(P(A\cap B)=P(A)\cdot P(B|A)=(2/3)(9/14)=3/7\).
Example 5. Three cards are drawn successively without replacement from a 52-card deck. Find the probability that the first two cards are kings and the third is an ace.
P(K₁) = 4/52. P(K₂|K₁) = 3/51. P(A₃|K₁∩K₂) = 4/50. Product = (4·3·4)/(52·51·50) = 48/132600 = 2/5525.
Example 6. A bag contains 5 red and 3 blue balls. A ball is drawn at random and put back; another ball is drawn. Find the probability that both are red.
With replacement, draws are independent. P(R₁) = 5/8, P(R₂) = 5/8. P(both red) = P(R₁)·P(R₂) = 25/64.
Example 7. P(E) = 0.3, P(F) = 0.4. Find P(E ∪ F) if E and F are (i) mutually exclusive, (ii) independent.
(i) Mutually exclusive: P(E∩F) = 0. P(E∪F) = 0.3 + 0.4 = 0.7.
(ii) Independent: P(E∩F) = 0.3·0.4 = 0.12. P(E∪F) = 0.3 + 0.4 − 0.12 = 0.58.
Activity: Card Probability with and without Replacement
L4 Analyse
Materials: Pen, paper, optional 52-card deck.
Predict: Two cards drawn from a deck. Compare P(both aces) WITH vs WITHOUT replacement.
  1. With replacement (independent draws): P(A₁)·P(A₂) = (4/52)·(4/52) = 1/169 ≈ 0.0059.
  2. Without replacement (conditional): P(A₁)·P(A₂|A₁) = (4/52)·(3/51) = 1/221 ≈ 0.0045.
  3. Without replacement gives slightly LOWER probability — taking one ace makes the second harder to draw.
  4. Now do "two cards, both same suit": with replacement = (1/4) (any first, then matching, ignoring first card's suit) — actually = (1/4); without = (13·12)/(52·51) = 156/2652 = 13/221 ≈ 0.0589 vs 1/4 = 0.25 (with replacement). Big difference: removing one card reshapes the deck.
  5. Lesson: replacement governs whether successive draws are independent.
"With replacement" produces independent trials — multiplication of unconditional probabilities. "Without replacement" requires the multiplication theorem with conditional probabilities at each step. This single distinction underlies the difference between the binomial (next chapter conceptually) and hypergeometric distributions.

Competency-Based Questions

Scenario: A factory has two production lines A and B. P(defective | A) = 0.05; P(defective | B) = 0.10. Line A produces 60% of items, B produces 40%.
Q1. P(item is from A AND defective):
L3 Apply
Answer: P(A) · P(defective | A) = 0.60 · 0.05 = 0.03.
Q2. P(item is defective) (overall):
L3 Apply
Answer: P(D) = P(A)P(D|A) + P(B)P(D|B) = 0.03 + 0.04 = 0.07. (This is total probability, formalised in Part 2.)
Q3. (T/F) "Two events with positive probabilities cannot be both mutually exclusive and independent." Justify.
L5 Evaluate
True. Mutually exclusive: P(E∩F)=0. Independent: P(E∩F)=P(E)P(F). For both: P(E)P(F)=0, requiring P(E)=0 or P(F)=0. Hence with positive probabilities, both can't hold.
Q4. A coin is biased: P(H) = 2/3. The coin is tossed 3 times independently. Find P(at least one head).
L4 Analyse
Solution: Use complement. P(no heads) = P(TTT) = (1/3)³ = 1/27. P(at least one head) = 1 − 1/27 = 26/27.
Q5. Design: a hospital test for a disease has sensitivity 99% (P(test+|disease)=0.99) and specificity 95% (P(test−|no disease)=0.95). Disease prevalence is 1%. Find P(disease | test+) — the question Bayes' theorem will answer in Part 2. Use only basic probability for now.
L6 Create
Solution: Out of 10000 people: 100 have disease (test+: 99). 9900 don't (test+: 5% = 495). Total test+ = 99 + 495 = 594. P(disease | test+) = 99/594 ≈ 0.167 — only 17%! Surprisingly low because false positives among the healthy outnumber true positives. This is the celebrated "base-rate fallacy" Bayes' theorem cleanly captures.

Assertion–Reason Questions

Assertion (A): P(E|F) is undefined when P(F) = 0.
Reason (R): Conditioning on an event of zero probability leads to division by zero.
(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 explains why A's restriction P(F)>0 is needed.
Assertion (A): If E and F are independent, then E' and F are also independent.
Reason (R): P(E'∩F) = P(F) − P(E∩F) = P(F) − P(E)P(F) = P(F)(1−P(E)) = P(F)P(E').
(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 algebraic proof of A.
Assertion (A): Mutually exclusive events with positive probabilities are independent.
Reason (R): Mutual exclusion means P(E∩F) = 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: (d). A is FALSE — mutual exclusion with positive probabilities precludes independence (since P(E)P(F) > 0 ≠ 0). R is true and is the precise reason A is false.

Frequently Asked Questions — Conditional Probability, Multiplication Theorem and Independent Events

What is conditional probability?
P(E|F) = P(E ∩ F)/P(F), the probability of E given F has occurred.
What is the multiplication theorem of probability?
P(E ∩ F) = P(E) · P(F|E) = P(F) · P(E|F).
What are independent events?
P(E ∩ F) = P(E)·P(F). Knowledge of one doesn't affect the other.
Difference between mutually exclusive and independent?
Mutually exclusive: cannot both occur (P(E∩F)=0). Independent: do not influence each other (P(E∩F)=P(E)P(F)).
How do you compute P(at least one) of n independent events?
P(at least one) = 1 − P(none) = 1 − (1−p₁)(1−p₂)…(1−pₙ).
Are E and E' independent?
No, in general. P(E∩E')=0 ≠ P(E)P(E') unless one has probability 0.
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