NCERT Solutions for Class 11 Maths Chapter 14 – Probability Miscellaneous Exercise
The Miscellaneous Exercise of Class 11 Maths Chapter 14 – Probability brings together all the key concepts covered in the chapter — classical probability, combination-based counting, the addition theorem, and complementary events. The problems in this exercise are more application-oriented and require students to think beyond direct formula application. Topics include drawing marbles, cards, and lottery tickets, forming numbers from digits, seating arrangements, and event-based probability using real-life scenarios.
This exercise is essential for CBSE board exam preparation as it tests conceptual clarity and problem-solving ability together. Questions involving combinations (nCr), the addition theorem, and the complement rule frequently appear in board exams and JEE. Mastering this exercise gives students a complete command over the entire Probability chapter and builds the analytical foundation needed for Class 12 probability topics.
NCERT Solutions for Class 11 Maths Chapter 14 – Probability Miscellaneous Exercise
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Question 1. A box contains 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that:
Answer:
Total marbles = 60 n(S) = ⁶⁰C₅
(i) All will be blue
5 blue marbles chosen from 20 blue marbles: n(A) = ²⁰C₅
P(all blue) = ²⁰C₅ / ⁶⁰C₅
(ii) At least one will be green
Using complement: P(at least one green) = 1 – P(no green)
Non-green marbles = 10 red + 20 blue = 30 P(no green) = ³⁰C₅ / ⁶⁰C₅
P(at least one green) = 1 – ³⁰C₅ / ⁶⁰C₅
Question 2. 4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining three diamonds and one spade?
Answer:
n(S) = ⁵²C₄
In a deck: 13 diamonds, 13 spades. Ways to draw 3 diamonds and 1 spade = ¹³C₃ × ¹³C₁
P(3 diamonds and 1 spade) = (¹³C₃ × ¹³C₁) / ⁵²C₄
Question 3. A die has two faces each with number '1', three faces each with number '2' and one face with number '3'. If the die is rolled once, determine:
Answer:
Total faces = 6
(i) P(2) Faces with 2 = 3 P(2) = 3/6 = 1/2
(ii) P(1 or 3) Faces with 1 = 2, faces with 3 = 1 P(1 or 3) = (2 + 1)/6 = 3/6 = 1/2
(iii) P(not 3) P(3) = 1/6 P(not 3) = 1 – 1/6 = 5/6
Question 4. 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:
Answer:
Total tickets = 10,000; Prizes = 10
(i) One ticket P(winning) = 10/10,000 = 1/1,000 P(not winning) = 1 – 1/1,000 = 999/1,000
(ii) Two tickets Non-prize tickets = 10,000 – 10 = 9,990 P(not winning) = ⁹'⁹⁹⁰C₂ / ¹⁰'⁰⁰⁰C₂
(iii) Ten tickets P(not winning) = ⁹'⁹⁹⁰C₁₀ / ¹⁰'⁰⁰⁰C₁₀
Question 5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that:
Answer:
Total ways to select 2 from 100 = ¹⁰⁰C₂
(i) You both enter the same section Ways both are in same section = ⁴⁰C₂ + ⁶⁰C₂
P(same section) = (⁴⁰C₂ + ⁶⁰C₂) / ¹⁰⁰C₂
= [(40×39)/2 + (60×59)/2] / [(100×99)/2]
= (780 + 1770) / 4950
= 2550/4950 = 17/33
(ii) You both enter different sections P(different sections) = 1 – 17/33 = 16/33
Question 6. Three letters are dictated to three persons and an envelope is addressed to each of them. Letters are inserted at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.
Answer:
Let letters be L₁, L₂, L₃ and envelopes be E₁, E₂, E₃. Total ways of inserting 3 letters in 3 envelopes = 3! = 6
All 6 arrangements:
- L₁E₁, L₂E₃, L₃E₂ → L₁ in correct envelope
- L₂E₂, L₁E₃, L₃E₁ → L₂ in correct envelope
- L₃E₃, L₁E₂, L₂E₁ → L₃ in correct envelope
- L₁E₁, L₂E₂, L₃E₃ → All three correct
- L₁E₂, L₂E₃, L₃E₁ → None correct
- L₁E₃, L₂E₁, L₃E₂ → None correct
Favorable cases (at least one correct) = 4
P(at least one letter in proper envelope) = 4/6 = 1/2
Question 7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find:
Answer:
(i) P(A ∪ B) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.54 + 0.69 – 0.35 = 0.88
(ii) P(A' ∩ B') P(A' ∩ B') = P(A ∪ B)' = 1 – P(A ∪ B) = 1 – 0.88 = 0.12
(iii) P(A ∩ B') P(A ∩ B') = P(A) – P(A ∩ B) = 0.54 – 0.35 = 0.19
(iv) P(B ∩ A') P(B ∩ A') = P(B) – P(A ∩ B) = 0.69 – 0.35 = 0.34
Question 8. From the employees of a company, 5 persons are selected. Details:
| S.No. | Name | Sex | Age |
|---|---|---|---|
| 1 | Harish | M | 30 |
| 2 | Rohan | M | 33 |
| 3 | Sheetal | F | 46 |
| 4 | Alis | F | 28 |
| 5 | Salim | M | 41 |
What is the probability that the spokesperson will be either male or over 35 years?
Answer:
n(S) = 5
Let E = event of selecting a male → E = {Harish, Rohan, Salim} → P(E) = 3/5 Let F = event of selecting person over 35 → F = {Sheetal, Salim} → P(F) = 2/5 E ∩ F = {Salim} (only male over 35) → P(E ∩ F) = 1/5
P(E ∪ F) = 3/5 + 2/5 – 1/5 = 4/5
Question 9. If four-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5 and 7, what is the probability of forming a number divisible by 5 when:
Answer:
(i) Digits are repeated Thousands digit: 5 or 7 (2 choices) Remaining 3 digits: 5 choices each → Total = 2 × 5 × 5 × 5 – 1 = 249 (excluding 5000)
Divisible by 5: unit digit must be 0 or 5 Total such numbers = (2 × 5 × 5 × 2) – 1 = 99 (excluding 5000)
P(divisible by 5) = 99/249 = 33/83
(ii) Repetition not allowed Total 4-digit numbers > 5000 = 2 × 4 × 3 × 2 = 48
When thousands digit = 5: unit place only 0 → 1 × 3 × 2 × 1 = 6 ways When thousands digit = 7: unit place 0 or 5 → 1 × 3 × 2 × 2 = 12 ways Total divisible by 5 = 6 + 12 = 18
P(divisible by 5) = 18/48 = 3/8
Question 10. The number lock of a suitcase has 4 wheels, each labelled with digits 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of getting the right sequence?
Answer:
Total 4-digit sequences without repetition = ¹⁰C₄ × 4! = 10 × 9 × 8 × 7 = 5040
Only 1 sequence opens the lock.
P(right sequence) = 1/5040
FAQs – Chapter 14 Probability Miscellaneous Exercise
Q1. How is combination used in probability problems?
When selecting items from a group where order does not matter, we use nCr = n! / (r! × (n–r)!). In Miscellaneous Exercise, questions on marbles, cards, and lottery tickets all use combinations to count favorable outcomes and total outcomes.
Q2. What is the complement rule and when is it used?
P(A') = 1 – P(A). It is most useful when calculating "at least one" type problems, where directly counting favorable cases is difficult. For example, P(at least one green) = 1 – P(no green).
Q3. How do you apply the addition theorem in this exercise?
P(A ∪ B) = P(A) + P(B) – P(A ∩ B). This is used in Questions 7 and 8 to find the probability of two overlapping events occurring together.
Q4. What does P(A ∩ B') mean?
P(A ∩ B') means the probability that event A occurs but event B does not. It equals P(A) – P(A ∩ B). This concept appears in Question 7(iii) and (iv).
Q5. Why is 5000 subtracted in Question 9?
Because the problem asks for four-digit numbers greater than 5,000. When the thousands digit is 5 and the remaining digits are all 0, it forms 5000 — which is not greater than 5,000. So it is excluded from the total count.