๐ CHAPTER 15: PROBABILITY
Subject: Mathematics | Class: XI | Kendriya Vidyalaya Sangathan
๐ INTRODUCTION
Probability is the measure of the likelihood that an event will occur. It is a number between 0 and 1 (inclusive).
๐น Key Concepts:
- Experiment: Any action with uncertain outcomes
- Sample Space (S): Set of all possible outcomes
- Event (E): A subset of the sample space
- Probability: P(E) = n(E)/n(S)
- Mutually Exclusive Events: Events that cannot occur together
- Exhaustive Events: Events whose union is the sample space
๐น Important Properties:
- 0 ≤ P(E) ≤ 1
- P(S) = 1
- P(ฯ) = 0
- P(E') = 1 - P(E)
- P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
๐ TYPES OF EVENTS
๐น Simple Event
An event with exactly one sample point.
๐น Compound Event
An event with more than one sample point.
๐น Mutually Exclusive Events
Two events A and B are mutually exclusive if they cannot occur simultaneously.
A ∩ B = ฯ
๐น Exhaustive Events
Events E₁, E₂, ..., Eโ are exhaustive if their union is the sample space.
E₁ ∪ E₂ ∪ ... ∪ Eโ = S
๐น Mutually Exclusive and Exhaustive Events
Events that are both mutually exclusive and exhaustive.
➕ ADDITION THEOREM OF PROBABILITY
๐น For Two Events:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
๐น For Mutually Exclusive Events:
P(A ∪ B) = P(A) + P(B)
๐น For Three Events:
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C)
๐น Complementary Events
P(A') = 1 - P(A)
P(A' ∩ B') = P(A ∪ B)' = 1 - P(A ∪ B)
๐ MCQ QUESTIONS
Q1. If (1+4p)/4, (1-p)/2, and (1-2p)/2 are the probabilities of three mutually exclusive events, then the value of p is:
(a) 1 (b) 1/2 (c) 1/3 (d) 2/3
Answer: (a) 1
Solution: Sum of probabilities = 1
(1+4p)/4 + (1-p)/2 + (1-2p)/2 = 1
(1+4p)/4 + (2-2p)/4 + (2-4p)/4 = 1
(5 - 2p)/4 = 1 ⇒ 5 - 2p = 4 ⇒ p = 1/2
Q2. Which of the following cannot be the probability of an event?
(a) 2/3 (b) -1/5 (c) 15% (d) 0.7
Answer: (b) -1/5
Reason: Probability cannot be negative.
Q3. Probability of an event can be:
(a) -0.7 (b) 11/9 (c) 1.001 (d) 0.6
Answer: (d) 0.6
Reason: Probability lies between 0 and 1.
Q4. In an experiment, the sum of probabilities of different events is:
(a) 1 (b) 0.5 (c) -2 (d) 0
Answer: (a) 1
Q5. In rolling a dice, the probability of getting number 8 is:
(a) 0 (b) 1 (c) -1 (d) 1/2
Answer: (a) 0
Q6. In a simultaneous throw of 2 coins, the probability of having 2 heads is:
(a) 1/4 (b) 1/2 (c) 1/8 (d) 1/6
Answer: (a) 1/4
Q7. The probability of getting sum more than 7 when a pair of dice are thrown is:
(a) 7/36 (b) 5/12 (c) 7/12 (d) None of these
Answer: (b) 5/12
Q8. The probability of raining on day 1 is 0.2 and on day 2 is 0.3. The probability of raining on both the days is:
(a) 0.2 (b) 0.1 (c) 0.06 (d) 0.25
Answer: (c) 0.06
Q9. If A and B are two events, such that P(A ∪ B) = 3/4, P(A ∩ B) = 1/4, P(A') = 2/3, where A' stands for the complementary event of A, then P(B) is given by:
(a) 1/9 (b) 2/3 (c) 1/3 (d) 2/9
Answer: (b) 2/3
Solution: P(A') = 2/3 ⇒ P(A) = 1/3
P(A∪B) = P(A) + P(B) - P(A∩B)
3/4 = 1/3 + P(B) - 1/4
P(B) = 3/4 - 1/3 + 1/4 = 1 - 1/3 = 2/3
Q10. The probability of getting at least one head when two coins are tossed is:
(a) 1/4 (b) 1/2 (c) 3/4 (d) 1
Answer: (c) 3/4
Solution: P(at least one head) = 1 - P(no head) = 1 - 1/4 = 3/4
Q11. A die is rolled. The probability of getting an even number is:
(a) 1/2 (b) 1/3 (c) 1/6 (d) 2/3
Answer: (a) 1/2
Q12. A card is drawn from a well-shuffled deck of 52 cards. The probability of getting a king is:
(a) 1/13 (b) 1/52 (c) 4/52 (d) Both A and C
Answer: (d) Both A and C
Q13. Two events A and B are mutually exclusive if:
(a) A ∩ B = ฯ (b) A ∪ B = S (c) A = B (d) None of these
Answer: (a) A ∩ B = ฯ
Q14. If A and B are mutually exclusive events, then P(A ∪ B) is:
(a) P(A) + P(B) (b) P(A) - P(B) (c) P(A) × P(B) (d) None of these
Answer: (a) P(A) + P(B)
Q15. If A and B are exhaustive events, then:
(a) A ∪ B = S (b) A ∩ B = ฯ (c) A = B (d) None of these
Answer: (a) A ∪ B = S
Q16. Probability of an impossible event is:
(a) 0 (b) 1 (c) 1/2 (d) -1
Answer: (a) 0
Q17. Probability of a certain event is:
(a) 0 (b) 1 (c) 1/2 (d) -1
Answer: (b) 1
Q18. In a single throw of a die, the probability of getting a number greater than 4 is:
(a) 1/6 (b) 1/3 (c) 2/3 (d) 1/2
Answer: (b) 1/3
Solution: Numbers > 4: 5, 6 ⇒ n(E) = 2, n(S) = 6 ⇒ P = 2/6 = 1/3
Q19. If an event has more than one sample point, then it is called a/an:
(a) Simple event (b) Elementary event (c) Compound event (d) None of these
Answer: (c) Compound event
Q20. A die is rolled. Let E be the event "die shows 4" and F be the event "die shows even number". Then, E and F are:
(a) Mutually exclusive (b) Exhaustive (c) Mutually exclusive and exhaustive (d) None of the above
Answer: (d) None of the above
Explanation: E ⊂ F, so they are not mutually exclusive.
๐ CASE BASED QUESTIONS
Q21. If three coins are tossed, all possible outcomes are {HHH, HHT, HTH, THH, TTH, HTT, THT, TTT}. If E₁, E₂, ..., Eโ are n events of a sample space S and if E₁ ∪ E₂ ∪ ... ∪ Eโ = S, then E₁, E₂, ..., Eโ are called exhaustive events. Further, if Eแตข ∩ Eโฑผ = ฯ for i ≠ j, then events E₁, E₂, ..., Eโ are called mutually exclusive and exhaustive events.
(i) Two events which are mutually exclusive.
Answer: "Getting at least two heads" and "Getting at least two tails"
(ii) Three events which are mutually exclusive and exhaustive.
Answer: "Getting no heads", "Getting exactly one head", "Getting at least two heads"
(iii) Two events, which are not mutually exclusive.
Answer: "Getting at most two tails" and "Getting exactly two tails"
Q22. Grandmother of Reena, Simi and Aryan has a bag, which contains 9 discs of different beautiful colors, of which 4 are red, 3 are blue and 2 are yellow. The discs are similar in shape and size. Children asked grandmother to give them discs and grandmother draws a disc at random from the bag.
(i) Calculate the probability that it will be red.
Answer: P(red) = 4/9
(ii) Calculate the probability that it will be yellow.
Answer: P(yellow) = 2/9
(iii) Calculate the probability that it will be blue.
Answer: P(blue) = 3/9 = 1/3
(iv) Calculate the probability that it will be not blue.
Answer: P(not blue) = 1 - 1/3 = 2/3
(v) Calculate the probability that it will be either red or yellow.
Answer: P(red or yellow) = (4+2)/9 = 6/9 = 2/3
Q23. An urn contains twenty white slips of paper numbered from 1 through 20, ten red slips of paper numbered from 1 through 10, forty yellow slips of paper numbered from 1 through 40 and ten blue slips of paper numbered from 1 through 10. These 80 slips of paper are thoroughly shuffled so that each slip has the equal chance of being drawn.
(i) What is the probability that slip drawn is blue or white?
Answer: P(blue or white) = (10+20)/80 = 30/80 = 3/8
(ii) What is the probability that slip drawn is numbered 1, 2, 3, 4 or 5?
Answer: Number of slips with 1-5: 4×5 = 20
P = 20/80 = 1/4
(iii) What is the probability that slip drawn is red or yellow and numbered 1, 2, 3 or 4?
Answer: Red: 4 slips, Yellow: 4 slips, Total: 8
P = 8/80 = 1/10
Q24. The given Venn diagram shows three events A, B and C and also the probabilities of the various intersections.
(i) The value of P(B ∩ C) is:
(a) 0.17 (b) 0.45 (c) 0.13 (d) 0.15
Answer: (a) 0.17
(ii) The value of P(A ∩ B) is:
(a) 0.17 (b) 0.45 (c) 0.13 (d) 0.15
Answer: (b) 0.45
(iii) The value of P(A ∩ C) is:
(a) 0.17 (b) 0.45 (c) 0.13 (d) 0.15
Answer: (c) 0.13
(iv) The value of P(B ∩ C') is:
(a) 0.17 (b) 0.45 (c) 0.13 (d) 0.15
Answer: (d) 0.15
Q25. The letters of the word 'ASSASSINATION' are arranged at random.
(i) The probability that four S's come consecutively in the word is:
(a) 2/143 (b) 2/143 (c) 25/26 (d) 15/26
Answer: (a) 2/143
(ii) The probability that two I's and two N's come together is:
(a) 2/143 (b) 2/143 (c) 25/26 (d) 15/26
Answer: (a) 2/143
(iii) The probability that all A's are not coming together is:
(a) 2/143 (b) 2/143 (c) 25/26 (d) 15/26
Answer: (c) 25/26
(iv) The probability that no two A's are coming together is:
(a) 2/143 (b) 2/143 (c) 25/26 (d) 15/26
Answer: (d) 15/26
(v) The probability that all vowels are coming together is:
(a) 2/143 (b) 2/143 (c) 25/26 (d) 15/26
Answer: (a) 2/143
๐ง ASSERTION & REASON QUESTIONS
Q26. Assertion: Probability of getting a head in a toss of an unbiased coin is 1/2.
Reason: In a simultaneous toss of two coins, the probability of getting 'no tails' is 1/4.
Answer: (b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.
Q27. Assertion: In tossing a coin, the exhaustive number of cases is 2.
Reason: If a pair of dice is thrown, then the exhaustive number of cases is 6 × 6 = 36.
Answer: (b) Both Assertion and Reason are correct.
Q28. Assertion: A letter is chosen at random from the word 'NAGATATION'. Then, the total number of outcomes is 10.
Reason: A letter is chosen at random from the word 'ASSASSINATION'. Then, the total number of outcomes is 13.
Answer: (b) Both Assertion and Reason are correct.
Q29. Consider a single throw of die and two events: A = the number is even = {2, 4, 6}, B = the number is a multiple of 3 = {3, 6}
Assertion: P(A ∪ B) = 2/3 and P(A ∩ B) = 1/6
Reason: P(A' ∩ B') = P(A ∪ B)' = 1 - 2/3 = 1/3
Answer: (b) Both Assertion and Reason are correct but Reason is not the correct explanation.
Q30. Assertion: Two dice are thrown simultaneously. There are 11 possible outcomes and each of them has a probability 1/11.
Reason: Probability of an event (E) is defined as P(E) = Number of favourable outcomes / Total number of outcomes.
Answer: (d) Assertion is incorrect, reason is correct.
Explanation: There are 36 outcomes, not 11. Sum outcomes are 2 through 12 (11 sums), but each sum has different probability.
๐ VERY SHORT ANSWER TYPE QUESTIONS
Q31. The numbers 1, 2, 3 and 4 are written separately on four slips of paper. The slips are put in a box and mixed thoroughly. A person draws two slips from the box, one after the other, without replacement. Describe the sample space for the experiment.
Answer: S = {(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)}
Q32. An experiment consists of rolling a die and then tossing a coin once if the number on the die is even. If the number on the die is odd, the coin is tossed twice. Write the sample space for this experiment.
Answer: S = {1HH, 1HT, 1TH, 1TT, 2H, 2T, 3HH, 3HT, 3TH, 3TT, 4H, 4T, 5HH, 5HT, 5TH, 5TT, 6H, 6T}
Q33. A box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
Answer: S = {RW, WR, WW}
Q34. If 3/4 is the probability of an event, what is the probability of the event "not A"?
Answer: P(A') = 1 - 3/4 = 1/4
Q35. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer: P(woman) = 6/10 = 3/5
Q36. If two cards are drawn from a well-shuffled pack, what is the probability that at least one of the two is heart?
Answer: P(at least one heart) = 1 - P(no heart) = 1 - (39/52)(38/51) = 1 - 741/2652 = 1911/2652
Q37. What is the probability that a leap year will have 53 Sundays?
Answer: 366 days = 52 weeks + 2 days
P(53 Sundays) = 2/7
Q38. What is the probability of getting a total of 10 in a single throw of two dice?
Answer: Outcomes for sum 10: (4,6), (5,5), (6,4) ⇒ 3 outcomes
P = 3/36 = 1/12
Q39. Two dice are rolled simultaneously. What is the probability that the numbers on them are different?
Answer: P(different) = 1 - P(same) = 1 - 6/36 = 30/36 = 5/6
Q40. At random all the letters of the word "ARTICLE" are arranged in all possible ways. What is the probability that the arrangement begins with vowel and ends with a consonant?
Answer: ARTICLE has 3 vowels (A, I, E) and 4 consonants (R, T, C, L)
First place: 3 choices, Last place: 4 choices
Remaining 5 letters: 5! ways
Total arrangements: 7!
P = (3×4×5!)/7! = (3×4×120)/5040 = 1440/5040 = 2/7
๐ SHORT ANSWER TYPE QUESTIONS
Q41. A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that letter is (i) a vowel (ii) a consonant (iii) Letter S.
Answer: Total letters = 13
(i) Vowels: A, A, I, A, I, O ⇒ 6 vowels ⇒ P = 6/13
(ii) Consonants: 7 ⇒ P = 7/13
(iii) Letter S: 4 times ⇒ P = 4/13
Q42. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B).
Answer: (i) P(A') = 1 - 0.42 = 0.58
(ii) P(B') = 1 - 0.48 = 0.52
(iii) P(A∪B) = 0.42 + 0.48 - 0.16 = 0.74
Q43. In Class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer: P(M) = 0.4, P(B) = 0.3, P(M∩B) = 0.1
P(M∪B) = 0.4 + 0.3 - 0.1 = 0.6
Q44. A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If die is rolled once, determine (i) P(2) (ii) P(1 or 3) (iii) P(not 3).
Answer: (i) P(2) = 3/6 = 1/2
(ii) P(1 or 3) = (2+1)/6 = 3/6 = 1/2
(iii) P(not 3) = 1 - 1/6 = 5/6
Q45. In a certain lottery, 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.
Answer: (a) 9990/10000 = 999/1000
(b) 9990C₂/10000C₂
(c) 9990C₁₀/10000C₁₀
Q46. A five digit number without repetition is formed by the digits 1, 2, 3, 4, 5, 6, 7, 8. What is the probability that the number has even digits at both ends?
Answer: Total numbers = ⁸P₅ = 6720
Even digits: 2,4,6,8 (4 choices for ends)
First and last: 4×3 = 12 ways
Middle 3 from remaining 6: ⁶P₃ = 120
Favourable = 12×120 = 1440
P = 1440/6720 = 3/14
Q47. 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?
Answer: P = (¹³C₃ × ¹³C₁)/⁵²C₄
Q48. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B).
Answer: (i) P(A') = 0.58
(ii) P(B') = 0.52
(iii) P(A∪B) = 0.74
Q49. In Class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer: 0.6
Q50. A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Answer: Possible amounts:
4 heads: +4 (P=1/16)
3 heads, 1 tail: 3 - 1.5 = +1.5 (P=4/16=1/4)
2 heads, 2 tails: 2 - 3 = -1 (P=6/16=3/8)
1 head, 3 tails: 1 - 4.5 = -3.5 (P=4/16=1/4)
0 heads, 4 tails: -6 (P=1/16)
✅ ANSWERS - MCQs
| Q.No. | Answer | Q.No. | Answer |
|---|---|---|---|
| 1 | (a) | 11 | (a) |
| 2 | (b) | 12 | (d) |
| 3 | (d) | 13 | (a) |
| 4 | (a) | 14 | (a) |
| 5 | (a) | 15 | (a) |
| 6 | (a) | 16 | (a) |
| 7 | (b) | 17 | (b) |
| 8 | (c) | 18 | (b) |
| 9 | (b) | 19 | (c) |
| 10 | (c) | 20 | (d) |
๐ KEY FORMULAS SUMMARY
| Formula | Expression |
|---|---|
| Probability of an Event | P(E) = n(E)/n(S) |
| Complementary Event | P(E') = 1 - P(E) |
| Addition Theorem (Two Events) | P(A∪B) = P(A) + P(B) - P(A∩B) |
| Addition Theorem (Mutually Exclusive) | P(A∪B) = P(A) + P(B) |
| Addition Theorem (Three Events) | P(A∪B∪C) = P(A)+P(B)+P(C) - P(A∩B) - P(B∩C) - P(C∩A) + P(A∩B∩C) |
| Mutually Exclusive | A ∩ B = ฯ |
| Exhaustive Events | A ∪ B = S |
๐น Important Points:
- 0 ≤ P(E) ≤ 1
- P(S) = 1
- P(ฯ) = 0
- Sum of probabilities of all elementary events = 1
๐ Kendriya Vidyalaya Sangathan | Class XI Mathematics
Chapter 15: Probability | Student Support Material