A die is rolled. Let E be the event 'die shows 4' and F be the event 'die shows even number'. Are E and F mutually exclusive?
Hint. Write out both events as sets and check whether their intersection is empty.
E = {4}, F = {2,4,6}. E intersection F = {4}, which is not empty.
✦ Working through each part gives: E and F are not mutually exclusive, since 4 is common to both.
A die is thrown. Describe the events: (i) A: a number less than 7 (ii) B: a number greater than 7 (iii) C: a multiple of 3 (iv) D: a number less than 4 (v) E: an even number greater than 4 (vi) F: a number not less than 3. Also find A union B, A intersection B, B union C, E intersection F, D intersection E, A-C, D-E, E intersection F', F'.
Hint. List each event as a subset of S={1,2,3,4,5,6}, then compute each set operation directly.
A={1,2,3,4,5,6}=S. B=empty set (impossible, no number exceeds 7 on a die). C={3,6}. D={1,2,3}. E={6}. F={3,4,5,6}. Then: A union B = S. A intersection B = empty set. B union C = {3,6}. E intersection F = {6}. D intersection E = empty set. A-C = {1,2,4,5}. D-E = {1,2,3}. F' = {1,2}, so E intersection F' = empty set.
✦ Working through each part gives: A=S, B=empty set, C={3,6}, D={1,2,3}, E={6}, F={3,4,5,6}; A union B=S, A intersection B=empty set, B union C={3,6}, E intersection F={6}, D intersection E=empty set, A-C={1,2,4,5}, D-E={1,2,3}, E intersection F'=empty set, F'={1,2}.
An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the events: A: the sum is greater than 8, B: 2 occurs on either die, C: the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive?
Hint. List each event, keeping in mind that a sum greater than 8 can never occur alongside a 2 on either die (since the other die would need to show more than 6).
A (sum>8) has 10 outcomes: sums 9,10,11,12. B (a 2 on either die) has 11 outcomes. C (sum at least 7 and a multiple of 3, i.e. sum=9 or 12) has 5 outcomes, and is actually a subset of A. Checking intersections: A intersection B is empty, since a sum over 8 with a 2 present would require the other die to show more than 6, which is impossible — so A and B are mutually exclusive. B intersection C is empty for the same reason (both C's cases, sum 9 or 12, are already impossible with a 2 present) — so B and C are mutually exclusive. A intersection C = C itself (nonempty, since C is a subset of A), so A and C are not mutually exclusive.
✦ Working through each part gives: A and B are mutually exclusive, and B and C are mutually exclusive; A and C are not mutually exclusive (C is a subset of A).
Three coins are tossed once. Let A denote the event 'three heads show', B denote 'two heads and one tail show', C denote 'three tails show', and D denote 'a head shows on the first coin'. Which events are (i) mutually exclusive? (ii) simple? (iii) compound?
Hint. List each event as a subset of S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}, then compare pairs for empty intersections, and count sample points in each event.
A={HHH}, B={HHT,HTH,THH}, C={TTT}, D={HHH,HHT,HTH,HTT}. Checking all pairs: A intersection B=empty, A intersection C=empty, B intersection C=empty, C intersection D=empty (since D requires a head first, C is all tails); but A intersection D={HHH} (nonempty, since A is a subset of D) and B intersection D={HHT,HTH} (nonempty). So the mutually exclusive pairs are (A,B), (A,C), (B,C), (C,D). For simple/compound: A and C each have exactly one sample point (simple events); B and D each have more than one (compound events).
✦ Working through each part gives: (i) Mutually exclusive pairs: (A,B), (A,C), (B,C), (C,D). (ii) Simple events: A and C. (iii) Compound events: B and D.
Three coins are tossed. Describe: (i) two events which are mutually exclusive (ii) three events which are mutually exclusive and exhaustive (iii) two events which are not mutually exclusive (iv) two events which are mutually exclusive but not exhaustive (v) three events which are mutually exclusive but not exhaustive.
Hint. Use S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT} and construct examples for each case directly; several valid answers exist for each part.
(i) 'All heads'={HHH} and 'all tails'={TTT} are disjoint, hence mutually exclusive. (ii) 'No head'={TTT}, 'exactly one head'={HTT,THT,TTH}, 'at least two heads'={HHT,HTH,THH,HHH}: these three sets are pairwise disjoint and their union is all of S, so mutually exclusive and exhaustive. (iii) 'At least one head' and 'at least one tail' overlap (e.g. on HHT), so they are not mutually exclusive. (iv) 'Exactly two heads'={HHT,HTH,THH} and 'exactly two tails'={HTT,THT,TTH} are disjoint (mutually exclusive) but their union misses HHH and TTT, so not exhaustive. (v) 'Exactly three heads'={HHH}, 'exactly three tails'={TTT}, and 'exactly two heads'={HHT,HTH,THH} are pairwise disjoint (mutually exclusive) but their union misses the 'exactly two tails' outcomes, so not exhaustive.
✦ Working through each part gives: See the worked examples above for one valid choice in each part; other correct answers are also possible.
Two dice are thrown. The events A, B, C are: A: getting an even number on the first die. B: getting an odd number on the first die. C: getting the sum of the numbers on the dice <= 5. Describe the events (i) A' (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) A intersection B' intersection C'.
Hint. Note first that A and B partition S exactly (every first-die value is either even or odd), so A'=B and B'=A. Then work out C explicitly (10 outcomes with sum<=5) and split it by whether the first die is even or odd.
A has 18 outcomes (first die even), B has 18 outcomes (first die odd), and A union B=S with A intersection B=empty, so A'=B and B'=A. C={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}, 10 outcomes; of these, 4 have an even first die ({(2,1),(2,2),(2,3),(4,1)}, so these are in A intersection C) and 6 have an odd first die (in B intersection C). (i) A'=B. (ii) not B = B'=A. (iii) A or B = A union B = S (the whole sample space). (iv) A and B = A intersection B = empty set. (v) A but not C = A-C, which removes the 4 even-first-die elements of C from A's 18, leaving 14 outcomes. (vi) B or C = B union C: since 6 of C's 10 elements are already in B, this adds the remaining 4 (even-first-die) elements, giving 18+4=22 outcomes. (vii) B and C = B intersection C, the 6 odd-first-die elements of C. (viii) A intersection B' intersection C' = A intersection A intersection C' = A intersection C' = A-C, the same 14-outcome set as part (v).
✦ Working through each part gives: (i) A'=B (18 outcomes). (ii) not B=A (18 outcomes). (iii) A or B=S (36 outcomes). (iv) A and B=empty set. (v) A but not C has 14 outcomes. (vi) B or C has 22 outcomes. (vii) B and C has 6 outcomes. (viii) Same 14-outcome set as (v).
Refer to question 6 above (the two-dice events A, B, C). State true or false, with reason: (i) A and B are mutually exclusive (ii) A and B are mutually exclusive and exhaustive (iii) A = B' (iv) A and C are mutually exclusive (v) A and B' are mutually exclusive (vi) A', B', C are mutually exclusive and exhaustive.
Hint. Reuse the sets and relationships established in question 6: A and B partition S, and A intersection C has 4 common elements.
(i) True, since A intersection B = empty set (a die's first value can't be both even and odd). (ii) True, since additionally A union B = S. (iii) True, since A and B partition S exactly, meaning each is the other's complement. (iv) False, since A intersection C has 4 elements (the even-first-die outcomes of C), not empty. (v) False, since B'=A, so A intersection B' = A intersection A = A, which is nonempty (in fact this is the same event twice over, maximally overlapping). (vi) False, since A'=B and B'=A, so this asks whether B, A, C are mutually exclusive and exhaustive — but A and C are not mutually exclusive (from part iv), so the whole statement fails.
✦ Working through each part gives: (i) True (ii) True (iii) True (iv) False (v) False (vi) False.
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.
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