Question 313 marksShort AnswerSection C

(a) Out of two bags, bag I contains 3 red and 4 white balls and bag II contains 8 red and 6 white balls. A die is thrown. If it shows a number less than 3 then a ball is drawn at random from bag I, otherwise a ball is drawn at random from bag II. Find the probability that the ball drawn from one of the bags is a red ball.


OR


(b) The probability of simultaneous occurrence of at least one of the two events X and Y is a. If the probability that exactly one of the events X, Y occurs is b, prove that P(X′)+P(Y′)=2−2a+bP(X') + P(Y') = 2 - 2a + b.

Probability
Probability — Total Probability Theorem and Complementary Events

Official answer

Option (a): P(red)=11/21P(\text{red}) = 11/21


  • A die showing a number less than 3 (1 or 2): P=2/6=1/3P = 2/6 = 1/3 → bag I; otherwise P=4/6=2/3P = 4/6 = 2/3 → bag II.
  • P(red∣bag I)=3/7P(\text{red} | \text{bag I}) = 3/7, P(red∣bag II)=8/14=4/7P(\text{red} | \text{bag II}) = 8/14 = 4/7.
  • By the Theorem of Total Probability: P(red)=(1/3)(3/7)+(2/3)(4/7)=1/7+8/21=P(\text{red}) = (1/3)(3/7) + (2/3)(4/7) = 1/7 + 8/21 = 11/21.

Option (b): P(X′)+P(Y′)=2−2a+bP(X') + P(Y') = 2 - 2a + b (proved)


  • Given P(X∪Y)=aP(X \cup Y) = a and P(exactly one of X,Y)=bP(\text{exactly one of } X, Y) = b.
  • P(X∪Y)=P(X)+P(Y)−P(X∩Y)=aP(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = a.
  • P(exactly one)=P(X)+P(Y)−2P(X∩Y)=bP(\text{exactly one}) = P(X) + P(Y) - 2P(X \cap Y) = b.
  • Subtracting: P(X∩Y)=a−bP(X \cap Y) = a - b, so P(X)+P(Y)=2a−bP(X) + P(Y) = 2a - b.
  • Therefore P(X′)+P(Y′)=[1−P(X)]+[1−P(Y)]=2−(2a−b)=P(X') + P(Y') = [1 - P(X)] + [1 - P(Y)] = 2 - (2a - b) = 2 − 2a + b.
Answer keywords:theorem of total probabilityconditional probabilityaddition theorem of probabilitycomplementary eventsprobability of exactly one eventbag and balls problem

Marking scheme

  1. 1.1 mark: correct P(bag I)=1/3P(\text{bag I}) = 1/3, P(bag II)=2/3P(\text{bag II}) = 2/3 and P(red∣bag I)=3/7P(\text{red}|\text{bag I}) = 3/7, P(red∣bag II)=4/7P(\text{red}|\text{bag II}) = 4/7 (a); OR correct equations P(X∪Y)=aP(X\cup Y) = a and P(exactly one)=bP(\text{exactly one}) = b (b).
  2. 2.1 mark: correct application of the Total Probability Theorem (a); OR correct derivation P(X∩Y)=a−bP(X\cap Y) = a - b and P(X)+P(Y)=2a−bP(X)+P(Y) = 2a - b (b).
  3. 3.1 mark: correct final answer 11/2111/21 (a); OR correct proof P(X′)+P(Y′)=2−2a+bP(X')+P(Y') = 2 - 2a + b (b).

Hint

For (a), use the Theorem of Total Probability with P(bag I)=1/3P(\text{bag I}) = 1/3, P(bag II)=2/3P(\text{bag II}) = 2/3. For (b), write P(X∪Y)=aP(X \cup Y) = a and P(exactly one)=bP(\text{exactly one}) = b in terms of P(X)P(X), P(Y)P(Y), P(X∩Y)P(X \cap Y), then use P(X′)=1−P(X)P(X') = 1 - P(X).

Quick oral answer

Part (a) uses the Theorem of Total Probability across two bags selected by a die roll, giving 11/2111/21; part (b) derives P(X)+P(Y)=2a−bP(X)+P(Y) = 2a - b, then takes complements to get P(X′)+P(Y′)=2−2a+bP(X')+P(Y') = 2 - 2a + b.

Common mistakes

  1. 1.Reversing which die outcomes correspond to bag I vs bag II (using ≥3\ge 3 for bag I instead of <3< 3).
  2. 2.Not reducing 8/148/14 to 4/74/7 before combining, leading to arithmetic errors in the final fraction.
  3. 3.In part (b), confusing P(exactly one)P(\text{exactly one}) with P(X∩Y)P(X \cap Y), or forgetting to apply the complement P(X′)=1−P(X)P(X') = 1 - P(X) at the end.

Analysis and explanation

This OR question tests the Theorem of Total Probability and algebraic manipulation of set-theoretic probability identities.


Concept: Part (a) treats the die roll as the event that selects which bag is used, then combines the two conditional probabilities of red using the Theorem of Total Probability. Part (b) expresses the two given quantities — a (probability of the union) and b (probability of exactly one) — in terms of P(X)P(X), P(Y)P(Y) and P(X∩Y)P(X \cap Y), solves for P(X) + P(Y), and finally uses complements.


Key identities for (b): P(X∪Y)=P(X)+P(Y)−P(X∩Y)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) and P(exactly one)=P(X)+P(Y)−2P(X∩Y)P(\text{exactly one}) = P(X) + P(Y) - 2P(X \cap Y). Subtracting isolates P(X∩Y)=a−bP(X \cap Y) = a - b.


Exam trap: In part (a), a common error is not reducing 8/14 to 4/7, or swapping which die outcomes map to which bag. In part (b), students confuse P(exactly one)P(\text{exactly one}) with P(X∩Y)P(X \cap Y), or forget the final complement step P(X′)=1−P(X)P(X') = 1 - P(X).

Interesting facts

The Theorem of Total Probability, central to part (a), is the foundation of Bayes' Theorem, published posthumously by Thomas Bayes in 1763 and now central to modern classifiers in machine learning.

Problems involving 'probability of exactly one event' are a direct application of the inclusion-exclusion principle, traceable to Abraham de Moivre's 1718 work on chance.

Question details

Q31 of 44 questions

Board
CBSE
Class
Class 12
Subject
Mathematics
Year
2026
Section
Section C
Marks
3
Type
Short Answer
Chapter
Probability
Topic
Probability — Total Probability Theorem and Complementary Events
Bloom's level
apply
Syllabus topic
Unit 6: Probability — Theorem of Total Probability and Probability of Complementary Events
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why use the Theorem of Total Probability in part (a)?

Because the ball is drawn from one of two mutually exclusive bags depending on the die outcome, the overall probability of red must be found by weighting each bag's conditional probability of red by the probability of that bag being chosen.

What is the key algebraic trick in proving part (b)?

Write both given probabilities — the union a and the exactly-one probability b — in terms of P(X)P(X), P(Y)P(Y) and P(X∩Y)P(X \cap Y); subtracting the two equations gives P(X∩Y)=a−bP(X \cap Y) = a - b, hence P(X)+P(Y)=2a−bP(X) + P(Y) = 2a - b, and finally P(X′)+P(Y′)=2−(2a−b)=2−2a+bP(X') + P(Y') = 2 - (2a - b) = 2 - 2a + b.