Miscellaneous Exercise — Probability

1. M.16 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A box contains 10 red, 20 blue, and 30 green marbles. 5 marbles are drawn from the box. What is the probability that (i) all will be blue? (ii) at least one will be green?

   Hint. The box has 60 marbles total; use combinations for the total and favourable ways, and use the complement for 'at least one green'.

   Total ways to draw 5 from 60: C(60,5)=5461512. (i) All blue: C(20,5)=15504 ways, so P=15504/5461512=34/11977. (ii) At least one green: use the complement 'no green', meaning all 5 come from the 10 red + 20 blue = 30 non-green marbles: C(30,5)=142506 ways, so P(no green)=142506/5461512, and P(at least one green)=1-142506/5461512.

   ✦ Working through each part gives: (i) P(all blue) = 34/11977 (approximately 0.0028). (ii) P(at least one green) is approximately 0.974.

2. M.24 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   4 cards are drawn from a well-shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

   Hint. Use combinations: choose 3 of the 13 diamonds and 1 of the 13 spades, divided by the total ways to choose any 4 cards from 52.

   Favourable ways = C(13,3) x C(13,1) = 286 x 13 = 3718. Total ways = C(52,4) = 270725.

   ✦ Working through each part gives: P(3 diamonds and 1 spade) = 3718/270725 = 286/20825.

3. M.34 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   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 (i) P(2) (ii) P(1 or 3) (iii) P(not 3).

   Hint. Since the six faces are not equally distributed among 1, 2, 3, find each probability as (number of matching faces)/6.

   P(1) = 2/6 = 1/3, P(2) = 3/6 = 1/2, P(3) = 1/6. (i) P(2) = 1/2. (ii) P(1 or 3) = P(1)+P(3) = 1/3+1/6 = 1/2 (these are mutually exclusive outcomes). (iii) P(not 3) = 1 - 1/6 = 5/6.

   ✦ Working through each part gives: (i) 1/2 (ii) 1/2 (iii) 5/6.

4. M.46 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   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?

   Hint. Since 9990 of the 10000 tickets are non-winning, use combinations to count the ways to pick only non-winning tickets out of the tickets bought.

   (a) P(no prize with 1 ticket) = 9990/10000 = 999/1000. (b) P(no prize with 2 tickets) = C(9990,2)/C(10000,2), which works out to approximately 0.998. (c) P(no prize with 10 tickets) = C(9990,10)/C(10000,10), which works out to approximately 0.990.

   ✦ Working through each part gives: (a) 999/1000 = 0.999 (b) approximately 0.998 (c) approximately 0.990.

5. M.55 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   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 (a) you both enter the same section? (b) you both enter the different sections?

   Hint. Fix your own section by chance, then find the probability your friend lands in the same section given one fewer spot remains in each section out of 99 total remaining students.

   P(both in the 40-section) = (40/100)(39/99), and P(both in the 60-section) = (60/100)(59/99). Adding these (mutually exclusive cases): P(same section) = 40(39)/(100)(99) + 60(59)/(100)(99) = (1560+3540)/9900 = 5100/9900 = 17/33.

   ✦ Working through each part gives: (a) P(same section) = 17/33 (b) P(different sections) = 1-17/33 = 16/33.

6. M.65 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   Three letters are dictated to three persons, and an envelope is addressed to each of them. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

   Hint. Use the complement: find the probability that no letter is in its correct envelope (a derangement of 3 objects), then subtract from 1.

   Total arrangements = 3! = 6. The number of derangements of 3 objects (no letter in its correct envelope) is 2 (for objects 1,2,3, the derangements are 2,3,1 and 3,1,2). So P(no letter correct) = 2/6 = 1/3.

   ✦ Working through each part gives: P(at least one letter in its proper envelope) = 1 - 1/3 = 2/3.

7. M.75 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A and B are two events such that P(A)=0.54, P(B)=0.69, and P(A intersection B)=0.35. Find (i) P(A union B) (ii) P(A' intersection B') (iii) P(A intersection B') (iv) P(B intersection A').

   Hint. Use the addition rule for (i), De Morgan's law for (ii), and A intersection B' = A - (A intersection B) style reasoning for (iii) and (iv).

   (i) P(A union B) = 0.54+0.69-0.35 = 0.88. (ii) P(A' intersection B') = P((A union B)') = 1-0.88 = 0.12. (iii) P(A intersection B') = P(A) - P(A intersection B) = 0.54-0.35 = 0.19. (iv) P(B intersection A') = P(B) - P(A intersection B) = 0.69-0.35 = 0.34.

   ✦ Working through each part gives: (i) 0.88 (ii) 0.12 (iii) 0.19 (iv) 0.34.

8. M.84 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   From the employees of a company, 5 persons are selected to represent them in the managing committee: Harish (M, 30), Rohan (M, 33), Sheetal (F, 46), Alis (F, 28), Salim (M, 41). A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

   Hint. List the 'male' event and the 'over 35' event as subsets of the 5 people, then apply the addition rule (noting any overlap).

   Male = {Harish, Rohan, Salim}, 3 people. Over 35 = {Sheetal(46), Salim(41)}, 2 people (Rohan at 33 does not qualify). The overlap (male AND over 35) = {Salim}, 1 person. Male or over 35 = 3+2-1 = 4 people out of 5.

   ✦ Working through each part gives: P(male or over 35) = 4/5.

9. M.96 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   If 4-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 (i) the digits are repeated? (ii) the repetition of digits is not allowed?

   Hint. The first digit must be 5 or 7 to exceed 5000; the last digit must be 0 or 5 for divisibility by 5; count total valid numbers and favourable ones separately for each case.

   (i) With repetition: first digit has 2 choices (5 or 7), the two middle digits have 5 choices each, giving 2x5x5=50 numbers per last-digit choice; total numbers = 2x5x5x5=250. For divisibility by 5, the last digit must be 0 or 5 (2 choices), giving 2x5x5x2=100 favourable numbers. P=100/250=2/5. (ii) Without repetition: first digit has 2 choices (5 or 7); for each, the remaining 3 digits are chosen without repetition from the remaining 4 digits, giving 4x3x2=24 arrangements per first-digit choice, so 2x24=48 total numbers. For divisibility, case first digit=5: last digit must be 0 (5 already used), remaining 2 middle digits from 3 choices: 3x2=6 numbers. Case first digit=7: last digit can be 0 or 5, and for each the 2 middle digits come from the remaining 3: 2x(3x2)=12 numbers. Total favourable = 6+12=18. P=18/48=3/8.

   ✦ Working through each part gives: (i) P = 2/5 (ii) P = 3/8.

10. M.103 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    The number lock of a suitcase has 4 wheels, each labelled with ten digits from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

    Hint. Count the total number of 4-digit sequences with no repeated digit from 10 available digits, since only one sequence is correct.

    Total sequences with no repetition = 10x9x8x7 = 5040. Only one of these is the correct sequence.

    ✦ Working through each part gives: P(correct sequence) = 1/5040.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh114.pdf) — Exercise 14.1 (7 questions), Exercise 14.2 (21 questions), plus the chapter's Miscellaneous Exercise (10 questions), 38 questions total. The old stub had swapped what Exercise 14.1 and 14.2 actually cover (describing 14.1 as being about sample spaces and 14.2 as being about event types, the reverse of the book's real structure) and invented a fabricated third exercise (14.3, 21 questions) to hold content that genuinely belongs to the real Exercise 14.2 — corrected, and the real 10-question Miscellaneous Exercise (which the old stub never mentioned) added. Confirmed against the CBSE curriculum PDF that 'random experiments; outcomes, sample space' is formative-only for this chapter, matching the book's own opening line ('We have studied about random experiment and sample space...') which recalls rather than re-teaches these ideas. Every combinatorial probability answer in this file (the marbles, cards, lottery, derangement, and 4-digit-number questions) was independently verified with a Python script (using exact fractions, not floating point) before being written up as a step-by-step solution, catching zero discrepancies against the hand-derived working.. Questions are referenced from the NCERT textbook for identification.