🎲 🃏 🎯
🎲 Chapter 11: Probability | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Conditional Probability, Bayes’ Theorem, Independent Events
✅ Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
🎲 Probability — This chapter covers mutually exclusive events, independent events, conditional probability, Bayes’ theorem, and probability using combinations. The notes have been updated according to the latest syllabus of 2080. Now you don’t need to go anywhere searching for the notes of this chapter because we are here to serve you.
📚 Exercise
📖 Basic Definitions
🎯 Mutually Exclusive Events
Two or more events are mutually exclusive if their simultaneous occurrence is not possible. Example: In a coin toss, head and tail cannot occur together.
Two or more events are mutually exclusive if their simultaneous occurrence is not possible. Example: In a coin toss, head and tail cannot occur together.
✅ Favourable Cases
Outcomes of a random experiment that entail the happening of an event. Example: In rolling a die, cases favourable to “getting an odd number” are 3.
Outcomes of a random experiment that entail the happening of an event. Example: In rolling a die, cases favourable to “getting an odd number” are 3.
🔄 Independent Events
Two events are independent if the occurrence of one does not affect the occurrence of the other. Example: Tossing a coin and rolling a die.
Two events are independent if the occurrence of one does not affect the occurrence of the other. Example: Tossing a coin and rolling a die.
⚠️ Dependent Events
Events where the outcome of one affects the probability of the other. Example: Drawing cards without replacement.
Events where the outcome of one affects the probability of the other. Example: Drawing cards without replacement.
🔢 Permutation and Combination Review
Permutation: Arrangement of objects in some order.
\[ P(n, r) = \frac{n!}{(n-r)!} \]
Combination: Selection of objects without regard to order.
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
\[ P(n, r) = \frac{n!}{(n-r)!} \]
Combination: Selection of objects without regard to order.
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
📐 Definition of Probability
If there are n exhaustive, mutually exclusive, and equally likely cases, and m of them are favourable to an event E, then:
\[
P(E) = \frac{m}{n} = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}
\]
\[
0 \le P(E) \le 1
\]
\[
P(\text{not } E) = 1 – P(E)
\]
📊 Conditional Probability
The probability of event A given that event B has occurred is:
\[
P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0
\]
Multiplication Theorem: \(P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)\)
For Independent Events: \(P(A \cap B) = P(A) \cdot P(B)\)
For Mutually Exclusive Events: \(P(A \cup B) = P(A) + P(B)\)
Multiplication Theorem: \(P(A \cap B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)\)
For Independent Events: \(P(A \cap B) = P(A) \cdot P(B)\)
For Mutually Exclusive Events: \(P(A \cup B) = P(A) + P(B)\)
🧠 Bayes’ Theorem
If \(E_1, E_2, \dots, E_n\) are mutually exclusive and exhaustive events, then:
\[
P(E_i|A) = \frac{P(E_i) \cdot P(A|E_i)}{\sum_{j=1}^{n} P(E_j) \cdot P(A|E_j)}
\]
📝 Solved Problems (15 Important Questions)
Q1 A coin is tossed three times. Find the probability of getting exactly two heads.
Solution: Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (n=8)
Favourable outcomes: HHT, HTH, THH (m=3)
\(P = \frac{3}{8}\)
Favourable outcomes: HHT, HTH, THH (m=3)
\(P = \frac{3}{8}\)
✅ \(P = \frac{3}{8}\)
Q2 Two dice are rolled. Find the probability that the sum is 7 or 11.
Solution: Total outcomes = 36
Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes
Sum 11: (5,6),(6,5) → 2 outcomes
Favourable = 6+2=8, \(P = \frac{8}{36} = \frac{2}{9}\)
Sum 7: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) → 6 outcomes
Sum 11: (5,6),(6,5) → 2 outcomes
Favourable = 6+2=8, \(P = \frac{8}{36} = \frac{2}{9}\)
✅ \(P = \frac{2}{9}\)
Q3 A card is drawn from a well-shuffled deck of 52 cards. Find the probability that it is a king or a red card.
Solution: P(King) = 4/52, P(Red) = 26/52, P(King ∩ Red) = 2/52
\(P(\text{King or Red}) = \frac{4}{52} + \frac{26}{52} – \frac{2}{52} = \frac{28}{52} = \frac{7}{13}\)
\(P(\text{King or Red}) = \frac{4}{52} + \frac{26}{52} – \frac{2}{52} = \frac{28}{52} = \frac{7}{13}\)
✅ \(P = \frac{7}{13}\)
Q4 From a bag containing 5 red and 3 green balls, two balls are drawn at random. Find the probability that both are red.
Solution: Total ways = \(C(8,2)=28\), Favourable = \(C(5,2)=10\)
\(P = \frac{10}{28} = \frac{5}{14}\)
\(P = \frac{10}{28} = \frac{5}{14}\)
✅ \(P = \frac{5}{14}\)
Q5 If P(A)=0.4, P(B)=0.5, and P(A∩B)=0.2, find P(A∪B) and P(A|B).
\(P(A \cup B) = P(A) + P(B) – P(A \cap B) = 0.4 + 0.5 – 0.2 = 0.7\)
\(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4\)
\(P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4\)
✅ \(P(A\cup B)=0.7, P(A|B)=0.4\)
Q6 A bag contains 4 white and 6 black balls. Two balls are drawn without replacement. Find the probability that both are white.
\(P(\text{both white}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}\)
✅ \(P = \frac{2}{15}\)
Q7 If A and B are independent with P(A)=0.3, P(B)=0.6, find P(A∩B) and P(A∪B).
\(P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.6 = 0.18\)
\(P(A \cup B) = 0.3 + 0.6 – 0.18 = 0.72\)
\(P(A \cup B) = 0.3 + 0.6 – 0.18 = 0.72\)
✅ \(P(A\cap B)=0.18, P(A\cup B)=0.72\)
Q8 In a class of 30 students, 18 play football, 15 play cricket, and 8 play both. Find the probability that a randomly selected student plays at least one game.
\(P(F) = \frac{18}{30}, P(C) = \frac{15}{30}, P(F \cap C) = \frac{8}{30}\)
\(P(F \cup C) = \frac{18}{30} + \frac{15}{30} – \frac{8}{30} = \frac{25}{30} = \frac{5}{6}\)
\(P(F \cup C) = \frac{18}{30} + \frac{15}{30} – \frac{8}{30} = \frac{25}{30} = \frac{5}{6}\)
✅ \(P = \frac{5}{6}\)
Q9 A die is rolled. Find the probability of getting a number greater than 2 given that it is an odd number.
Odd numbers: {1,3,5}, numbers >2 from odd: {3,5}
\(P = \frac{2}{3}\)
\(P = \frac{2}{3}\)
✅ \(P = \frac{2}{3}\)
Q10 Two cards are drawn without replacement from a deck. Find the probability that both are aces.
\(P = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}\)
✅ \(P = \frac{1}{221}\)
Q11 If P(A)=0.6, P(B)=0.7, and P(A∪B)=0.9, are A and B independent?
\(P(A \cap B) = P(A)+P(B)-P(A \cup B) = 0.6+0.7-0.9 = 0.4\)
\(P(A) \cdot P(B) = 0.6 \times 0.7 = 0.42 \neq 0.4\)
So not independent.
\(P(A) \cdot P(B) = 0.6 \times 0.7 = 0.42 \neq 0.4\)
So not independent.
✅ Not independent
Q12 A box contains 5 red, 4 blue, and 3 green balls. Three balls are drawn at random. Find the probability that all are of different colors.
Total ways = \(C(12,3)=220\)
Favourable = \(5 \times 4 \times 3 = 60\)
\(P = \frac{60}{220} = \frac{3}{11}\)
Favourable = \(5 \times 4 \times 3 = 60\)
\(P = \frac{60}{220} = \frac{3}{11}\)
✅ \(P = \frac{3}{11}\)
Q13 Bayes’ Theorem: A factory has three machines A, B, C producing 30%, 40%, 30% of total output. Defective rates: 2%, 3%, 4%. Find the probability a defective item came from machine B.
P(A)=0.3, P(B)=0.4, P(C)=0.3
P(D|A)=0.02, P(D|B)=0.03, P(D|C)=0.04
P(B|D) = \(\frac{0.4 \times 0.03}{0.3\times0.02 + 0.4\times0.03 + 0.3\times0.04} = \frac{0.012}{0.006+0.012+0.012} = \frac{0.012}{0.03} = 0.4\)
P(D|A)=0.02, P(D|B)=0.03, P(D|C)=0.04
P(B|D) = \(\frac{0.4 \times 0.03}{0.3\times0.02 + 0.4\times0.03 + 0.3\times0.04} = \frac{0.012}{0.006+0.012+0.012} = \frac{0.012}{0.03} = 0.4\)
✅ \(P(B|D) = 0.4\)
Q14 A coin is tossed. If it shows head, a die is rolled; if tail, a card is drawn from a deck. Find the probability of getting a tail and a king.
P(Tail) = 1/2, P(King|Tail) = 4/52 = 1/13
\(P(\text{Tail and King}) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}\)
\(P(\text{Tail and King}) = \frac{1}{2} \times \frac{1}{13} = \frac{1}{26}\)
✅ \(P = \frac{1}{26}\)
Q15 Three coins are tossed. Find the probability of getting at least two heads.
Sample space: 8 outcomes. Favourable: HHT, HTH, THH, HHH (4 outcomes)
\(P = \frac{4}{8} = \frac{1}{2}\)
\(P = \frac{4}{8} = \frac{1}{2}\)
✅ \(P = \frac{1}{2}\)
✏️ Practice Questions
P1 Four coins are tossed. Find the probability of getting exactly three heads.
✅ Answer: \(\frac{4}{16} = \frac{1}{4}\)
P2 Two dice are rolled. Find the probability that the sum is 8.
✅ Answer: \(\frac{5}{36}\)
P3 From a bag of 7 red and 5 blue balls, two balls are drawn without replacement. Find the probability that both are blue.
✅ Answer: \(\frac{5}{33}\)
P4 If P(A)=0.5, P(B)=0.4, and P(A∩B)=0.2, find P(A|B).
✅ Answer: 0.5
P5 A card is drawn from a deck. Find the probability that it is a spade or a queen.
✅ Answer: \(\frac{4}{13}\)
📋 Multiple Choice Questions (HSEB Pattern)
MCQ 1 The probability of an event always lies between:
A) -1 and 0 B) 0 and 1 C) -1 and 1 D) 0 and 100
✅ Answer: B) 0 and 1
MCQ 2 If A and B are mutually exclusive, then P(A∪B) =
A) P(A) + P(B) B) P(A) – P(B) C) P(A)×P(B) D) P(A)/P(B)
✅ Answer: A) P(A) + P(B)
MCQ 3 If A and B are independent, then P(A∩B) =
A) P(A) + P(B) B) P(A) – P(B) C) P(A)×P(B) D) P(A)/P(B)
✅ Answer: C) P(A)×P(B)
MCQ 4 The probability of getting a head in a single coin toss is:
A) 1 B) 0 C) 1/2 D) 1/3
✅ Answer: C) 1/2
MCQ 5 The conditional probability formula is:
A) P(A|B)=P(A∩B)/P(B) B) P(A|B)=P(A∪B)/P(B) C) P(A|B)=P(B)/P(A) D) None
✅ Answer: A) P(A|B)=P(A∩B)/P(B)
MCQ 6 Two events are said to be mutually exclusive if:
A) They can occur together B) They cannot occur together C) They are independent D) None
✅ Answer: B) They cannot occur together
MCQ 7 If a die is rolled, what is the probability of getting an even number?
A) 1/6 B) 1/3 C) 1/2 D) 2/3
✅ Answer: C) 1/2
MCQ 8 Bayes’ theorem is used to find:
A) Prior probability B) Posterior probability C) Marginal probability D) None
✅ Answer: B) Posterior probability
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This PDF provides the solutions of every question from the 1st exercise of class 12 probability chapter. Since this chapter has only one exercise, you don’t have to search for other exercises.
This PDF provides the solutions of every question from the 1st exercise of class 12 probability chapter. Since this chapter has only one exercise, you don’t have to search for other exercises.
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📌 Key Formulas Summary:
• Probability: \(P(E) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}\)
• Addition Rule: \(P(A \cup B) = P(A) + P(B) – P(A \cap B)\)
• Conditional Probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
• Independent Events: \(P(A \cap B) = P(A) \cdot P(B)\)
• Multiplication Theorem: \(P(A \cap B) = P(A) \cdot P(B|A)\)
• Bayes’ Theorem: \(P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)}\)
• Probability: \(P(E) = \frac{\text{Favourable outcomes}}{\text{Total outcomes}}\)
• Addition Rule: \(P(A \cup B) = P(A) + P(B) – P(A \cap B)\)
• Conditional Probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)
• Independent Events: \(P(A \cap B) = P(A) \cdot P(B)\)
• Multiplication Theorem: \(P(A \cap B) = P(A) \cdot P(B|A)\)
• Bayes’ Theorem: \(P(E_i|A) = \frac{P(E_i)P(A|E_i)}{\sum P(E_j)P(A|E_j)}\)