Exercise 14.1
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?
Answer
E = {4}, F = {2, 4, 6}
E ∩ F = {4} ≠ ∅
Since E and F have a common outcome (4), they are not mutually exclusive.
A die is thrown. Describe the events A–F and find the required set operations.
Answer
Sample space S = {1, 2, 3, 4, 5, 6}
(i) A: a number less than 7 ⟹ A = {1, 2, 3, 4, 5, 6}
(ii) B: a number greater than 7 ⟹ B = {} = ∅
(iii) C: a multiple of 3 ⟹ C = {3, 6}
(iv) D: a number less than 4 ⟹ D = {1, 2, 3}
(v) E: an even number greater than 4 ⟹ E = {6}
(vi) F: a number not less than 3 ⟹ F = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}; A ∩ B = ∅
B ∪ C = {3, 6}; E ∩ F = {6}
D ∩ E = ∅; A – C = {1, 2, 4, 5}
D – E = {1, 2, 3}; E ∩ F′ = {6} ∩ {1, 2} = ∅
F′ = {1, 2}
Rolling a pair of dice. Describe events A, B, C and determine which pairs are mutually exclusive.
Answer
A: sum > 8 ⟹ A = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}
B: 2 occurs on either die ⟹ B = {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2)}
C: sum ≥ 7 and multiple of 3 ⟹ sum = 9 or 12 ⟹ C = {(3,6),(4,5),(5,4),(6,3),(6,6)}
A ∩ B = ∅ (no outcome with 2 on a die gives sum > 8) ⟹ A and B are mutually exclusive
B ∩ C = ∅ ⟹ B and C are mutually exclusive
A ∩ C = {(3,6),(4,5),(5,4),(6,3),(6,6)} ≠ ∅ ⟹ A and C are NOT mutually exclusive
Three coins are tossed. Events: A = three heads, B = two heads and one tail, C = three tails, D = head shows on first coin.
Answer
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
A = {HHH}, B = {HHT, HTH, THH}, C = {TTT}, D = {HHH, HHT, HTH, HTT}
(i) Mutually exclusive: A and B (A∩B=∅), A and C (A∩C=∅), B and C (B∩C=∅), A and C, C and D. A, B, C are pairwise mutually exclusive.
(ii) Simple events (single outcome): A = {HHH} and C = {TTT}
(iii) Compound events: B (3 outcomes) and D (4 outcomes)
Three coins are tossed. Describe the required events.
Answer
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two mutually exclusive events:
E₁ = {HHH, HHT, HTH, HTT} (head on first coin), E₂ = {THH, THT, TTH, TTT} (tail on first coin) — E₁∩E₂=∅
(ii) Three mutually exclusive and exhaustive events:
E₁={HHH}, E₂={HHT,HTH,THH,HTT,THT,TTH}, E₃={TTT}
(iii) Two events which are not mutually exclusive:
E₁ = {HHH, HHT, HTH, HTT} and E₂ = {HHH, THH, HTH, TTH} — both contain HHH
(iv) Two mutually exclusive but not exhaustive:
E₁={HHH}, E₂={TTT} — these don't cover the entire sample space
(v) Three mutually exclusive but not exhaustive:
E₁={HHH}, E₂={HHT}, E₃={HTH}
Two dice are thrown. A: even on first die, B: odd on first die, C: sum ≤ 5. Describe the required events.
Answer
A = {(2,1),(2,2),...,(2,6),(4,1),...,(4,6),(6,1),...,(6,6)} — all (even, any)
B = {(1,1),(1,2),...,(1,6),(3,1),...,(3,6),(5,1),...,(5,6)} — all (odd, any)
C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}
(i) A′ = B (getting odd on first die)
(ii) not B = A (getting even on first die)
(iii) A or B = A ∪ B = S (the entire sample space, since first die is even or odd)
(iv) A and B = A ∩ B = ∅ (a number cannot be both even and odd)
(v) A but not C = A ∩ C′: outcomes in A but NOT in C = {(2,4),(2,5),(2,6),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6),...}
(vi) B or C = B ∪ C = B ∪ {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}
Since B contains all (odd, *), the extra from C is {(2,1),(2,2),(2,3),(4,1)} already covered by C but not B.
B ∪ C = B ∪ {(2,1),(2,2),(2,3),(4,1)}
(vii) B and C = B ∩ C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}
(viii) A ∩ B′ ∩ C′ = A ∩ A ∩ C′ = A ∩ C′ = outcomes in A not in C (same as A but not C above)
State true or false (refer to Q6 events):
Answer
(i) A and B are mutually exclusive — TRUE (A∩B=∅)
(ii) A and B are mutually exclusive and exhaustive — TRUE (A∪B=S, A∩B=∅)
(iii) A = B′ — TRUE (complement of odd first die = even first die)
(iv) A and C are mutually exclusive — FALSE (both contain (2,1),(2,2),(2,3),(4,1))
(v) A and B′ are mutually exclusive — A and B′ = A and A = A; A∩A≠∅ so FALSE
(vi) A′, B′, C are mutually exclusive and exhaustive — A′=B, B′=A; B∩A=∅ but B∪A=S, and C is a subset of S but B∩C≠∅ so they are NOT mutually exclusive — FALSE
Exercise 14.2
Which assignments of probabilities are valid for S = {ω₁,...,ω₇}?
Answer
(a) 0.1, 0.01, 0.05, 0.03, 0.01, 0.2, 0.6: Sum = 1.00 ✓; all ≥ 0 ✓ — Valid
(b) 1/7 each: Sum = 7×(1/7) = 1 ✓; all ≥ 0 ✓ — Valid
(c) 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7: Sum = 2.8 ≠ 1 — Not valid
(d) –0.1, 0.2, 0.3, 0.4, –0.2, 0.1, 0.3: Contains negative values — Not valid
(e) 1/14, 2/14, 3/14, 4/14, 5/14, 6/14, 15/14: 15/14 > 1 and sum ≠ 1 — Not valid
A coin is tossed twice. What is the probability that at least one tail occurs?
Answer
S = {HH, HT, TH, TT}; n(S) = 4
Event (at least one tail) = {HT, TH, TT}; n(E) = 3
P(E) = 3/4 = 0.75
A die is thrown. Find the probability of the following events:
Answer
S = {1, 2, 3, 4, 5, 6}
(i) Prime number: {2, 3, 5}; P = 3/6 = 1/2
(ii) Number ≥ 3: {3, 4, 5, 6}; P = 4/6 = 2/3
(iii) Number ≤ 1: {1}; P = 1/6
(iv) Number > 6: {}; P = 0
(v) Number < 6: {1, 2, 3, 4, 5}; P = 5/6
A card is selected from a pack of 52 cards.
Answer
(a) Number of points in sample space = 52
(b) P(ace of spades) = 1/52
(c)(i) P(ace) = 4/52 = 1/13
(c)(ii) P(black card) = 26/52 = 1/2
A fair coin (1 on one face, 6 on the other) and a fair die are tossed. Find P(sum = 3) and P(sum = 12).
Answer
Coin shows 1 or 6; die shows 1–6. Total outcomes = 2×6 = 12.
(i) Sum = 3: coin=1, die=2 ⟹ (1,2); P = 1/12
(ii) Sum = 12: coin=6, die=6 ⟹ (6,6); P = 1/12
Four men and six women on city council. One selected at random. How likely is it to be a woman?
Answer
P(woman) = 6/10 = 3/5
A fair coin tossed four times. Win ₹1 for each head, lose ₹1.50 for each tail. Find different amounts possible and their probabilities.
Answer
Possible amounts (h heads, 4–h tails):
h=0: 0(1)–4(1.5) = –₹6; h=1: ₹1–₹4.5 = –₹3.5; h=2: ₹2–₹3 = –₹1; h=3: ₹3–₹1.5 = ₹1.5; h=4: ₹4–₹0 = ₹4
Total sample points = 2⁴ = 16
P(h=0) = C(4,0)/16 = 1/16; P(h=1) = C(4,1)/16 = 4/16 = 1/4
P(h=2) = C(4,2)/16 = 6/16 = 3/8; P(h=3) = C(4,3)/16 = 4/16 = 1/4; P(h=4) = 1/16
Five different amounts: –₹6 (P=1/16), –₹3.5 (P=1/4), –₹1 (P=3/8), ₹1.5 (P=1/4), ₹4 (P=1/16)
Three coins tossed once. Find the probability of:
Answer
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}; n(S) = 8
(i) 3 heads: {HHH}; P = 1/8
(ii) 2 heads: {HHT, HTH, THH}; P = 3/8
(iii) At least 2 heads: {HHH, HHT, HTH, THH}; P = 4/8 = 1/2
(iv) At most 2 heads: S – {HHH}; P = 7/8
(v) No head: {TTT}; P = 1/8
(vi) 3 tails: {TTT}; P = 1/8
(vii) Exactly two tails: {HTT, THT, TTH}; P = 3/8
(viii) No tail: {HHH}; P = 1/8
(ix) At most two tails: S – {TTT}; P = 7/8
If P(A) = 2/11, what is P(not A)?
Answer
P(not A) = 1 – P(A) = 1 – 2/11 = 9/11
A letter chosen at random from 'ASSASSINATION'. Find P(vowel) and P(consonant).
Answer
ASSASSINATION has 13 letters: A-S-S-A-S-S-I-N-A-T-I-O-N
Vowels (A, I, O): A appears 3 times, I appears 2 times, O appears 1 time = 6 vowels
Consonants: 13–6 = 7 consonants
(i) P(vowel) = 6/13
(ii) P(consonant) = 7/13
In a lottery, a person chooses 6 different numbers from 1 to 20. What is the probability of winning?
Answer
Total ways to choose 6 from 20 = C(20,6) = 20!/(6!×14!) = 38760
Only one way to win.
P(winning) = 1/38760 ≈ 2.58 × 10⁻⁵
Check whether the following probabilities are consistently defined:
Answer
(i) P(A)=0.5, P(B)=0.7, P(A∩B)=0.6:
P(A∩B) must satisfy P(A∩B) ≤ P(A). But 0.6 > 0.5. Not consistently defined.
(ii) P(A)=0.5, P(B)=0.4, P(A∪B)=0.8:
P(A∪B) = P(A)+P(B)–P(A∩B) ⟹ 0.8 = 0.5+0.4–P(A∩B) ⟹ P(A∩B) = 0.1
P(A∩B)=0.1 ≥ 0 and ≤ min(0.5, 0.4) = 0.4 ✓. Consistently defined.
Fill in the blanks:
Answer
P(A∪B) = P(A)+P(B)–P(A∩B)
(i) P(A)=1/3, P(B)=1/5, P(A∩B)=1/15: P(A∪B) = 1/3+1/5–1/15 = 5/15+3/15–1/15 = 7/15
(ii) P(A)=0.35, P(A∩B)=0.25, P(A∪B)=0.6: P(B) = 0.6–0.35+0.25 = 0.5
(iii) P(A)=0.5, P(B)=0.35, P(A∪B)=0.7: P(A∩B) = 0.5+0.35–0.7 = 0.15
Given P(A) = 3/5 and P(B) = 1/5. Find P(A or B) if A and B are mutually exclusive.
Answer
If A and B are mutually exclusive, P(A∩B) = 0.
P(A∪B) = P(A)+P(B) = 3/5+1/5 = 4/5
E and F are events with P(E)=1/4, P(F)=1/2, P(E and F)=1/8. Find (i) P(E or F) and (ii) P(not E and not F).
Answer
(i) P(E∪F) = P(E)+P(F)–P(E∩F) = 1/4+1/2–1/8 = 2/8+4/8–1/8 = 5/8
(ii) P(E′∩F′) = P((E∪F)′) = 1–P(E∪F) = 1–5/8 = 3/8
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
Answer
P(E′∪F′) = 0.25 ⟹ P((E∩F)′) = 0.25 ⟹ P(E∩F) = 1–0.25 = 0.75
Since P(E∩F) = 0.75 ≠ 0, E and F are not mutually exclusive.
P(A)=0.42, P(B)=0.48, P(A and B)=0.16. Determine (i) P(not A), (ii) P(not B), (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
In Class XI, 40% study Mathematics, 30% study Biology, 10% study both. Find P(student studies 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
P(pass 1st exam) = 0.8, P(pass 2nd exam) = 0.7, P(pass at least one) = 0.95. Find P(pass both).
Answer
P(A∪B) = P(A)+P(B)–P(A∩B)
0.95 = 0.8+0.7–P(A∩B)
P(A∩B) = 1.5–0.95 = 0.55
P(pass both English and Hindi) = 0.5, P(pass neither) = 0.1, P(pass English) = 0.75. Find P(pass Hindi).
Answer
Let E = English, H = Hindi.
P(E∩H) = 0.5, P(E′∩H′) = 0.1 ⟹ P(E∪H) = 1–0.1 = 0.9
P(E∪H) = P(E)+P(H)–P(E∩H)
0.9 = 0.75+P(H)–0.5
P(H) = 0.9–0.75+0.5 = 0.65
In a class of 60 students: 30 opted NCC, 32 opted NSS, 24 opted both. If one is selected randomly, find:
Answer
P(NCC) = 30/60 = 1/2; P(NSS) = 32/60; P(NCC∩NSS) = 24/60 = 2/5
(i) P(NCC or NSS) = P(NCC)+P(NSS)–P(NCC∩NSS) = 30/60+32/60–24/60 = 38/60 = 19/30
(ii) P(neither) = 1–19/30 = 11/30
(iii) P(NSS but not NCC) = P(NSS)–P(NCC∩NSS) = 32/60–24/60 = 8/60 = 2/15
Miscellaneous Exercise
A box contains 10 red, 20 blue and 30 green marbles. 5 drawn. Find P(all blue) and P(at least one green).
Answer
Total marbles = 60; choosing 5: C(60,5) = 5461512
(i) P(all blue) = C(20,5)/C(60,5) = 15504/5461512 = 4/1403 ≈ 0.00285
(ii) P(at least one green) = 1 – P(no green) = 1 – C(30,5)/C(60,5)
= 1 – 142506/5461512 = 1 – 0.0261 ≈ 0.974
4 cards drawn from a deck of 52. Find P(3 diamonds and 1 spade).
Answer
C(13,3) × C(13,1) / C(52,4) = 286×13/270725 = 3718/270725 = 143/10425 ≈ 0.0137
A die has two faces with '1', three faces with '2', one face with '3'. Rolled once. Find P(2), P(1 or 3), P(not 3).
Answer
(i) P(2) = 3/6 = 1/2
(ii) P(1 or 3) = P(1)+P(3) = 2/6+1/6 = 3/6 = 1/2
(iii) P(not 3) = 1–1/6 = 5/6
10,000 tickets sold; 10 prizes. Find P(not winning) if you buy (a) 1 ticket, (b) 2 tickets, (c) 10 tickets.
Answer
(a) P(not winning) = (10000–10)/10000 = 9990/10000 = 999/1000
(b) P(not winning) = C(9990,2)/C(10000,2) = (9990×9989)/(10000×9999) = 99800110/99990000 ≈ 0.998
(c) P(not winning) = C(9990,10)/C(10000,10) ≈ 0.990
Two sections of 40 and 60 formed from 100 students. If you and your friend are among 100 students, find:
Answer
(a) P(both in same section):
P(both in section of 40) = C(98,38)/C(100,40) = (40×39)/(100×99) = 1560/9900 = 26/165
P(both in section of 60) = C(98,58)/C(100,60) = (60×59)/(100×99) = 3540/9900 = 59/165
P(same section) = 26/165+59/165 = 85/165 = 17/33
(b) P(different sections) = 1–17/33 = 16/33
Three letters to three persons; envelopes addressed to each; inserted randomly. Find P(at least one in proper envelope).
Answer
Total arrangements = 3! = 6
No letter in proper envelope (derangements for n=3) = 2
P(no letter correct) = 2/6 = 1/3
P(at least one correct) = 1–1/3 = 2/3
P(A)=0.54, P(B)=0.69, P(A∩B)=0.35. Find (i) P(A∪B), (ii) P(A′∩B′), (iii) P(A∩B′), (iv) P(B∩A′).
Answer
(i) P(A∪B) = 0.54+0.69–0.35 = 0.88
(ii) P(A′∩B′) = P((A∪B)′) = 1–0.88 = 0.12
(iii) P(A∩B′) = P(A)–P(A∩B) = 0.54–0.35 = 0.19
(iv) P(B∩A′) = P(B)–P(A∩B) = 0.69–0.35 = 0.34
Five persons: Harish(M,30), Rohan(M,33), Sheetal(F,46), Alis(F,28), Salim(M,41). One selected as spokesperson. Find P(male or over 35).
Answer
Male: {Harish, Rohan, Salim} = 3 persons
Over 35: {Sheetal(46), Salim(41)} = 2 persons
Male AND over 35: {Salim} = 1 person
P(male or over 35) = (3+2–1)/5 = 4/5 = 0.8
4-digit numbers > 5000 from digits 0, 1, 3, 5, 7. Find P(divisible by 5) when (i) digits repeated, (ii) no repetition.
Answer
(i) Digits repeated: First digit must be 5 or 7 (for >5000). Total = 2×5³ = 250.
Divisible by 5: last digit = 0 or 5. First digit = 5 or 7.
Last digit = 0: first = 5 or 7, middle 2 = 5 each ⟹ 2×5×5×1 = 50
Last digit = 5, first = 5: 1×5×5×1 = 25; first = 7: 1×5×5×1 = 25 ⟹ 50
Total favorable = 100; P = 100/250 = 2/5
(ii) No repetition: Digits 0,1,3,5,7. First digit = 5 or 7.
Total = C(1,1)×P(4,3)+C(1,1)×P(4,3)... First=5: remaining 3 from {0,1,3,7} = 4×3×2=24; first=7: remaining 3 from {0,1,3,5} but >5000 already ⟹ 4×3×2=24. But subtract cases where first=5 and number=5000... Total 4-digit > 5000 = 24+24=48.
Divisible by 5: last digit=0 or 5.
Last=0, first=5: remaining 2 from {1,3,7} = 3×2=6; last=0, first=7: remaining 2 from {1,3,5}=3×2=6; last=5, first=7: remaining 2 from {0,1,3}=3×2=6. (Last=5, first=5 not allowed since repetition not allowed.)
Favorable = 6+6+6=18; P = 18/48 = 3/8
A number lock has 4 wheels with digits 0–9. Opens with a 4-digit sequence with no repeats. Find P(getting right sequence).
Answer
Total sequences of 4 digits from 10, no repeats = P(10,4) = 10×9×8×7 = 5040
Only 1 correct sequence.
P = 1/5040 ≈ 1.98 × 10⁻⁴
Important Extra Questions
If A and B are mutually exclusive events with P(A) = 0.35 and P(B) = 0.45, find P(A∪B), P(A∩B), P(A′∩B′).
Answer
Since A and B are mutually exclusive: P(A∩B) = 0
P(A∪B) = 0.35+0.45 = 0.80
P(A′∩B′) = P((A∪B)′) = 1–0.80 = 0.20
A card is drawn from a deck of 52. Find the probability that it is a king or a red card.
Answer
P(king) = 4/52; P(red) = 26/52; P(red king) = 2/52
P(king or red) = 4/52+26/52–2/52 = 28/52 = 7/13
Two dice are thrown. Find the probability that the sum is 7 or 11.
Answer
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
P(sum = 7 or 11) = (6+2)/36 = 8/36 = 2/9
In a group of 1000 people, 540 can speak Hindi only, 320 can speak English only, and 140 can speak both. A person is selected at random. Find P(speaks Hindi), P(speaks English), P(speaks both or neither).
Answer
P(Hindi) = (540+140)/1000 = 680/1000 = 0.68
P(English) = (320+140)/1000 = 460/1000 = 0.46
P(both) = 140/1000 = 0.14
P(neither) = 1–P(H∪E) = 1–(0.68+0.46–0.14) = 1–1.00 = 0 (everyone speaks at least one)
P(both or neither) = 0.14+0 = 0.14
The probability that a student passes Physics is 2/3, passes Mathematics is 4/9. The probability of passing at least one of them is 4/5. Find the probability that the student passes both.
Answer
P(P∪M) = P(P)+P(M)–P(P∩M)
4/5 = 2/3+4/9–P(P∩M)
P(P∩M) = 2/3+4/9–4/5 = 30/45+20/45–36/45 = 14/45