Question 1
A die is rolled. E is “the die shows 4” and F is “the die shows an even number”. Are E and F mutually exclusive?
Steps only
E={4}; F={2,4,6}.
Their intersection is {4}, not empty.
Answer
No. E∩F={4}, so they are not mutually exclusive.
CLASS 11 · MATHEMATICS · CHAPTER 14
Probability
Describe events with sets, check probability assignments and solve equally likely and counting problems. Short answers come first, followed by clear reasoning.
41 worked answersAll 2 exercise setsAll 10 end-of-chapter questions
What you will learn
- Represent events as subsets of a sample space.
- Use unions, intersections and complements.
- Apply probability axioms and inclusion-exclusion.
- Count outcomes with permutations and combinations.
41 worked answers · short steps first, explanations when you need them
Exercise 14.1
Textbook pages 295–296
Question 2
For one die roll, describe A: less than 7; B: greater than 7; C: multiple of 3; D: less than 4; E: even and greater than 4; F: not less than 3. Find the listed unions/intersections/differences and complements.
Steps only
The die sample space is S={1,2,3,4,5,6}.
Translate each phrase into a subset of S.
Use union, intersection, difference and complement definitions.
Answer
A={1,2,3,4,5,6}; B=∅; C={3,6}; D={1,2,3}; E={6}; F={3,4,5,6}. A∪B=A; A∩B=∅; B∪C=C; E∩F={6}; D∩E=∅; A−C={1,2,4,5}; D−E=D; E∩F′=∅; F′={1,2}.
Question 3
Two dice are rolled. A is sum>8, B means a 2 appears on either die, and C means sum≥7 and divisible by 3. Describe the events and identify mutually exclusive pairs.
Steps only
A∩C consists of sums 9 or 12, so it is non-empty.
If a die shows 2, the sum is at most 8; hence B∩A=B∩C=∅.
Answer
A: sums 9–12; B: at least one die is 2; C: sums 9 or 12. A and B are mutually exclusive; B and C are mutually exclusive; A and C are not.
Question 4
Three coins are tossed. A=three heads, B=two heads and one tail, C=three tails, D=head on the first coin. Identify mutually exclusive pairs, simple events and compound events.
Steps only
A={HHH}; B={HHT,HTH,THH}; C={TTT}.
D={HHH,HHT,HTH,HTT}.
Simple events contain one outcome; compound events contain several.
Answer
Mutually exclusive pairs: A–B, A–C, B–C, C–D. Simple: A,C. Compound: B,D.
Question 5
For three coin tosses, give examples of (i) two mutually exclusive events, (ii) three mutually exclusive and exhaustive events, (iii) overlapping events, (iv) mutually exclusive but not exhaustive events, and (v) three mutually exclusive but not exhaustive events.
Steps only
The sample space has eight outcomes.
For disjoint events, check that no outcome is shared.
For exhaustive events, check that their union is all eight outcomes.
Answer
Examples:
- (i) first coin H / first coin T;
- (ii) 0 heads / exactly 1 head / at least 2 heads;
- (iii) at least 1 head / first coin H;
- (iv) HHH / TTT;
- (v) HHH / HHT / HTH.
Question 6
Two dice are thrown. A=even on first die, B=odd on first die, C=sum≤5. Describe A′, not B, A or B, A and B, A but not C, B or C, B and C, and A∩B′∩C′.
Steps only
A and B partition outcomes by the parity of the first die.
C contains ordered pairs whose sum is 2,3,4 or 5.
Apply the set operations to these outcome conditions.
Answer
A′=B; B′=A; A∪B=S; A∩B=∅; A−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)}; B∪C={first die odd}∪{sum≤5}; B∩C={(1,1),(1,2),(1,3),(1,4),(3,1),(3,2),(5,1)}; A∩B′∩C′=A−C.
Question 7
Using the events in Question 6, state true or false: (i) A,B mutually exclusive; (ii) mutually exclusive and exhaustive; (iii) A=B′; (iv) A,C mutually exclusive; (v) A,B′ mutually exclusive; (vi) A′,B′,C mutually exclusive and exhaustive.
Steps only
A and B are opposite first-die parities, so they are disjoint and cover S.
A=B′ because B is odd on the first die.
A∩C is non-empty, and A∩B′=A.
Answer
- (i) True;
- (ii) True;
- (iii) True;
- (iv) False;
- (v) False;
- (vi) False.
Exercise 14.2
Textbook pages 306–308
Question 1
Which assignments of probabilities to seven outcomes are invalid? (a) 0.1,0.01,0.05,0.03,0.01,0.2,0.6; (b) seven copies of 1/7; (c) 0.1 through 0.7; (d) −0.1,0.2,0.3,0.4,−0.2,0.1,0.3; (e) 1/14,2/14,3/14,4/14,5/14,6/14,15/14.
Steps only
(a) entries are non-negative and sum to 1; (b) entries sum to 1.
(c) sums to 2.8.
(d) has negative entries; (e) has 15/14>1.
Answer
Invalid: (c), (d), and (e).
Question 2
A coin is tossed twice. Find the probability of at least one tail.
Steps only
Outcomes: HH, HT, TH, TT.
Three of four outcomes contain a tail.
Answer
3/4.
Question 3
A die is thrown. Find probabilities of: (i) prime; (ii) ≥3; (iii) ≤1; (iv) >6; (v) <6.
Steps only
S={1,2,3,4,5,6}.
Favourable counts: 3,4,1,0,5.
Divide each count by 6.
Answer
- (i) 1/2;
- (ii) 2/3;
- (iii) 1/6;
- (iv) 0;
- (v) 5/6.
Question 4
One card is selected from a standard 52-card deck. Find sample-space size, P(ace of spades), P(ace), and P(black card).
Steps only
There are 52 distinct cards.
One ace of spades; four aces; 26 black cards.
Divide each count by 52.
Answer
52; 1/52; 1/13; 1/2.
Question 5
A fair coin marked 1 and 6 is tossed with a fair die. Find the probability the sum is (i) 3, (ii) 12.
Steps only
There are 2×6=12 equally likely pairs.
Sum 3 occurs for coin 1 and die 2.
Sum 12 occurs for coin 6 and die 6.
Answer
Both probabilities are 1/12.
Question 6
A city council has 4 men and 6 women. One member is selected at random. Find the probability of a woman.
Steps only
Total members=10.
P(woman)=6/10=3/5.
Answer
3/5.
Question 7
A fair coin is tossed four times. Win ₹1 per head and lose ₹1.50 per tail. Find each possible net amount and its probability.
Steps only
With h heads, amount=h−1.5(4−h)=2.5h−6.
For h=0,1,2,3,4 the outcome counts are 1,4,6,4,1.
Divide by 16.
Answer
Amounts: −₹6 (1/16), −₹3.50 (1/4), −₹1 (3/8), ₹1.50 (1/4), ₹4 (1/16).
Question 8
For three coins, find probabilities of (i) 3 heads; (ii) 2 heads; (iii) at least 2 heads; (iv) at most 2 heads; (v) no head; (vi) 3 tails; (vii) exactly 2 tails; (viii) no tail; (ix) at most 2 tails.
Steps only
There are 8 equally likely outcomes.
Counts for 0,1,2,3 heads are 1,3,3,1.
Use the corresponding counts; tails follow by symmetry.
Answer
- (i) 1/8;
- (ii) 3/8;
- (iii) 1/2;
- (iv) 7/8;
- (v) 1/8;
- (vi) 1/8;
- (vii) 3/8;
- (viii) 1/8;
- (ix) 7/8.
Question 9
If P(A)=2/11, find the probability of not A.
Steps only
P(A′)=1−P(A)=1−2/11.
Answer
9/11.
Question 10
A letter is chosen at random from ASSASSINATION. Find probabilities of vowel and consonant.
Steps only
The word has 13 letters.
Vowels: A occurs 3 times, I twice, O once: 6 total.
The other 7 letters are consonants.
Answer
Vowel=6/13; consonant=7/13.
Question 11
Six different numbers are chosen from 1 to 20. Find the probability they match a fixed winning set of six numbers.
Steps only
There are C(20,6)=38,760 unordered sets.
Exactly one set matches the winning numbers.
Answer
1/38,760.
Question 12
Check consistency: (i) P(A)=0.5,P(B)=0.7,P(A∩B)=0.6; (ii) P(A)=0.5,P(B)=0.4,P(A∪B)=0.8.
Steps only
For (i), an intersection cannot exceed P(A)=0.5.
For (ii), P(A∩B)=0.5+0.4−0.8=0.1, which is allowable.
Answer
- (i) Inconsistent;
- (ii) consistent, with P(A∩B)=0.1.
Question 13
Fill the missing values: (i) P(A)=1/3,P(B)=1/5,P(A∩B)=1/15; (ii) P(A)=0.35,P(A∩B)=0.25,P(A∪B)=0.6; (iii) P(A)=0.5,P(B)=0.35,P(A∪B)=0.7.
Steps only
Use P(A∪B)=P(A)+P(B)−P(A∩B).
(i) 1/3+1/5−1/15=7/15.
Rearrange the same rule for (ii) and (iii).
Answer
- (i) P(A∪B)=7/15;
- (ii) P(B)=0.5;
- (iii) P(A∩B)=0.15.
Question 14
If P(A)=3/5 and P(B)=1/5 and A,B are mutually exclusive, find P(A or B).
Steps only
Mutually exclusive means P(A∩B)=0.
P(A∪B)=3/5+1/5=4/5.
Answer
4/5.
Question 15
If P(E)=1/4,P(F)=1/2,P(E∩F)=1/8, find P(E∪F) and P(E′∩F′).
Steps only
P(E∪F)=1/4+1/2−1/8=5/8.
By De Morgan, P(E′∩F′)=1−P(E∪F)=3/8.
Answer
P(E∪F)=5/8; P(E′∩F′)=3/8.
Question 16
If P(E′∪F′)=0.25, can E and F be mutually exclusive?
Steps only
By De Morgan, E′∪F′=(E∩F)′.
Thus P(E∩F)=1−0.25=0.75.
Answer
No. P(E∩F)=0.75, so they overlap.
Question 17
If P(A)=0.42,P(B)=0.48,P(A∩B)=0.16, find P(A′),P(B′),P(A∪B).
Steps only
P(A′)=1−0.42=0.58.
P(B′)=1−0.48=0.52.
P(A∪B)=0.42+0.48−0.16=0.74.
Answer
0.58, 0.52, 0.74.
Question 18
40% of a class studies Mathematics, 30% Biology and 10% both. Find probability a student studies Mathematics or Biology.
Steps only
P(M∪B)=0.40+0.30−0.10.
=0.60.
Answer
0.60.
Question 19
P(pass first exam)=0.8, P(pass second)=0.7, P(pass at least one)=0.95. Find probability of passing both.
Steps only
P(A∪B)=P(A)+P(B)−P(A∩B).
P(A∩B)=0.8+0.7−0.95=0.55.
Answer
0.55.
Question 20
P(pass both English and Hindi)=0.5, P(pass neither)=0.1, P(pass English)=0.75. Find P(pass Hindi).
Steps only
P(E∪H)=1−0.1=0.9.
0.9=0.75+P(H)−0.5.
P(H)=0.65.
Answer
0.65.
Question 21
In a class of 60, 30 chose NCC, 32 chose NSS, 24 chose both. Find probabilities of (i) NCC or NSS, (ii) neither, (iii) NSS but not NCC.
Steps only
Union count=30+32−24=38.
Neither=60−38=22.
NSS only=32−24=8. Divide each count by 60.
Answer
- (i) 19/30;
- (ii) 11/30;
- (iii) 2/15.
Miscellaneous Exercise on Chapter 14
Textbook pages 310–312
Question 1(i)
A box has 10 red, 20 blue and 30 green marbles. Five are drawn without replacement. Find the probability all five are blue.
Steps only
Choose 5 from the 20 blue marbles.
Divide by the number of ways to choose any 5 of 60.
Answer
C(20,5)/C(60,5).
Question 1(ii)
For the same box and draw, find probability of at least one green marble.
Steps only
Use the complement: no green means all five are among the 30 red or blue marbles.
Subtract that probability from 1.
Answer
1−C(30,5)/C(60,5).
Question 2
Four cards are drawn from a shuffled 52-card deck. Find probability of exactly 3 diamonds and 1 spade.
Steps only
Choose 3 of the 13 diamonds and 1 of the 13 spades.
Total four-card hands=C(52,4).
Probability=286/20,825.
Answer
C(13,3)C(13,1)/C(52,4)=286/20,825.
Question 3
A die has two faces marked 1, three faces marked 2 and one face marked 3. Find P(2), P(1 or 3), P(not 3).
Steps only
There are six equally likely faces.
Counts are 3, 2+1, and 5.
Divide by 6.
Answer
1/2, 1/2, 5/6.
Question 4
A lottery sells 10,000 tickets and awards 10 prizes. Find probability of no prize when buying (a) one ticket, (b) two, (c) ten.
Steps only
There are 9,990 non-winning tickets.
For k tickets, choose all k from those 9,990.
Divide by C(10,000,k).
Answer
(a) 999/1000; (b) C(9990,2)/C(10000,2); (c) C(9990,10)/C(10000,10).
Question 5(a)
Two of 100 students are assigned to sections of 40 and 60. Find probability you and your friend enter the same section.
Steps only
P(same)=(40/100)(39/99)+(60/100)(59/99).
=(1560+3540)/9900=17/33.
Answer
17/33.
Question 5(b)
For the same assignment, find probability you and your friend enter different sections.
Steps only
Different sections is the complement of same section.
1−17/33=16/33.
Answer
16/33.
Question 6
Three letters are placed randomly into their three addressed envelopes. Find probability at least one letter is in its correct envelope.
Steps only
There are 3!=6 assignments.
Only 2 assignments have no letter in its correct envelope.
P(at least one correct)=1−2/6=2/3.
Answer
2/3.
Question 7
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′).
Steps only
Union=0.54+0.69−0.35=0.88.
Neither=1−0.88=0.12.
A only=0.54−0.35; B only=0.69−0.35.
Answer
- (i) 0.88;
- (ii) 0.12;
- (iii) 0.19;
- (iv) 0.34.
Question 8
From Harish (M,30), Rohan (M,33), Sheetal (F,46), Alis (F,28), Salim (M,41), choose a spokesperson. Find probability the person is male or over 35.
Steps only
Male set: Harish,Rohan,Salim.
Over-35 set: Sheetal,Salim.
Union has 4 people out of 5.
Answer
4/5.
Question 9(i)
Form four-digit numbers greater than 5000 from 0,1,3,5,7 with repetition allowed. Find probability divisible by 5.
Steps only
Total eligible: first digit 5 or 7 gives 250, but exclude 5000, so 249.
Favourable: first digit 5 gives 49 after excluding 5000; first digit 7 gives 50.
Probability=99/249=33/83.
Answer
33/83.
Question 9(ii)
For the same number-forming task, if repetition is not allowed, find probability divisible by 5.
Steps only
Total: first digit 5 or 7; arrange 3 of remaining 4 digits: 2×4P3=48.
Favourable ending 0 or 5: 6 numbers starting 5 and 12 starting 7.
18/48=3/8.
Answer
3/8.
Question 10
A four-wheel lock uses digits 0–9 with no repeats. What is the probability of guessing its exact sequence?
Steps only
Possible codes=10×9×8×7=5040.
Exactly one opens the lock.
Answer
1/5040.