Exercise 14.2 — Probability

1. 14.2.16 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   Which of the following can not be a valid assignment of probabilities for outcomes of sample space S={w1,w2,w3,w4,w5,w6,w7}? (a) 0.1,0.01,0.05,0.03,0.01,0.2,0.6 (b) 1/7 each (c) 0.1,0.2,0.3,0.4,0.5,0.6,0.7 (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.

   Hint. Check two things for each assignment: every value must lie in [0,1], and all seven values must sum to exactly 1.

   (a) Sum = 0.1+0.01+0.05+0.03+0.01+0.2+0.6 = 1, all values in [0,1]: valid. (b) Sum = 7(1/7) = 1, all values in [0,1]: valid. (c) Sum = 0.1+0.2+0.3+0.4+0.5+0.6+0.7 = 2.8, not equal to 1: not valid. (d) Contains negative values (-0.1 and -0.2), violating P(wi)>=0: not valid. (e) The last value 15/14 exceeds 1, and the sum is 36/14, not equal to 1: not valid.

   ✦ Working through each part gives: (a) and (b) are valid; (c), (d), and (e) are not valid.

2. 14.2.22 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A coin is tossed twice, what is the probability that at least one tail occurs?

   Hint. List the sample space and find how many outcomes have at least one tail; or use the complement (no tails at all).

   S={HH,HT,TH,TT}, n(S)=4. Only HH has no tail, so 'at least one tail' has the other 3 outcomes.

   ✦ Working through each part gives: P(at least one tail) = 3/4.

3. 14.2.35 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A die is thrown, find the probability of the following events: (i) A prime number will appear (ii) A number greater than or equal to 3 will appear (iii) A number less than or equal to 1 will appear (iv) A number more than 6 will appear (v) A number less than 6 will appear.

   Hint. List each favourable set out of S={1,2,3,4,5,6} and divide by 6.

   (i) Primes: {2,3,5}, 3 outcomes. (ii) >=3: {3,4,5,6}, 4 outcomes. (iii) <=1: {1}, 1 outcome. (iv) >6: impossible, 0 outcomes. (v) <6: {1,2,3,4,5}, 5 outcomes.

   ✦ Working through each part gives: (i) 3/6=1/2 (ii) 4/6=2/3 (iii) 1/6 (iv) 0 (v) 5/6.

4. 14.2.45 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A card is selected from a pack of 52 cards. (a) How many points are there in the sample space? (b) Calculate the probability that the card is an ace of spades. (c) Calculate the probability that the card is (i) an ace (ii) a black card.

   Hint. The sample space has one point per card; count aces and black cards directly out of 52.

   (a) n(S)=52. (b) There is exactly one ace of spades, so P = 1/52. (c)(i) There are 4 aces, so P=4/52=1/13. (ii) Half the deck (spades and clubs) is black, so 26 black cards, P=26/52=1/2.

   ✦ Working through each part gives: (a) 52 (b) 1/52 (c)(i) 1/13 (ii) 1/2.

5. 14.2.54 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A fair coin with 1 marked on one face and 6 on the other, and a fair die, are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12.

   Hint. List all 12 equally likely (coin, die) pairs and find which give each target sum.

   There are 2x6=12 equally likely outcomes. (i) Sum=3: only (coin=1, die=2) works, since (coin=6, die=-3) is impossible. That's 1 outcome. (ii) Sum=12: only (coin=6, die=6) works, 1 outcome.

   ✦ Working through each part gives: (i) P(sum=3) = 1/12 (ii) P(sum=12) = 1/12.

6. 14.2.63 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

   Hint. Divide the number of women by the total council size.

   Total council members = 4+6=10. Favourable (women) = 6.

   ✦ Working through each part gives: P(woman) = 6/10 = 3/5.

7. 14.2.76 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses, and the probability of having each of these amounts.

   Hint. For k heads out of 4 tosses, the net amount is k(1) + (4-k)(-1.50) = 2.5k-6; work out this value and its probability (using the binomial count of ways to get k heads) for k=0,1,2,3,4.

   With k heads (and 4-k tails), the amount is 2.5k-6. For k=0: amount=-6, 1 way (TTTT), probability 1/16. For k=1: amount=-3.5, 4 ways, probability 4/16=1/4. For k=2: amount=-1, 6 ways, probability 6/16=3/8. For k=3: amount=1.5, 4 ways, probability 4/16=1/4. For k=4: amount=4, 1 way (HHHH), probability 1/16.

   ✦ Working through each part gives: 5 different amounts: -Rs 6 (P=1/16), -Rs 3.50 (P=1/4), -Re 1 (P=3/8), Rs 1.50 (P=1/4), Rs 4 (P=1/16).

8. 14.2.89 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) at least 2 heads (iv) at most 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) at most 2 tails.

   Hint. Use S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}, n(S)=8, and count favourable outcomes for each part directly.

   (i) {HHH}: 1/8. (ii) {HHT,HTH,THH}: 3/8. (iii) 2 or 3 heads: 3+1=4 outcomes, 4/8=1/2. (iv) all except HHH: 7/8. (v) {TTT}: 1/8. (vi) {TTT}, same as no head: 1/8. (vii) {HTT,THT,TTH}: 3/8. (viii) {HHH}, same as no tail meaning all heads: 1/8. (ix) all except TTT: 7/8.

   ✦ Working through each part gives: (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.

9. 14.2.92 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

   If 2/11 is the probability of an event, what is the probability of the event 'not A'?

   Hint. Apply the complement rule directly.

   P(not A) = 1 - P(A) = 1 - 2/11.

   ✦ Working through each part gives: P(not A) = 9/11.

10. 14.2.104 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    A letter is chosen at random from the word 'ASSASSINATION'. Find the probability that the letter is (i) a vowel (ii) a consonant.

    Hint. Count the total letters and how many are vowels versus consonants.

    ASSASSINATION has 13 letters: A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, T appears once, O appears once (3+4+2+2+1+1=13). Vowels are A, I, O: 3+2+1=6. Consonants are S, N, T: 4+2+1=7.

    ✦ Working through each part gives: (i) P(vowel) = 6/13 (ii) P(consonant) = 7/13.

11. 14.2.114 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?

    Hint. Since order doesn't matter, count the total number of ways to choose 6 numbers from 20, and note only one of these matches the fixed set.

    The total number of ways to choose 6 numbers from 20 (order irrelevant) is C(20,6) = 38760. Only one such combination matches the committee's fixed numbers.

    ✦ Working through each part gives: P(winning) = 1/38760.

12. 14.2.124 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    Check whether the following probabilities P(A) and P(B) are consistently defined: (i) P(A)=0.5, P(B)=0.7, P(A intersection B)=0.6 (ii) P(A)=0.5, P(B)=0.4, P(A union B)=0.8.

    Hint. Use the fact that P(A intersection B) can never exceed the smaller of P(A) and P(B), and that P(A union B) must lie between max(P(A),P(B)) and P(A)+P(B).

    (i) P(A intersection B) can be at most min(P(A),P(B))=min(0.5,0.7)=0.5, but 0.6 exceeds this, so this assignment is not consistent. (ii) Computing the implied intersection: P(A intersection B)=P(A)+P(B)-P(A union B)=0.5+0.4-0.8=0.1, which lies between 0 and min(0.5,0.4)=0.4, so this assignment is consistent.

    ✦ Working through each part gives: (i) Not consistently defined. (ii) Consistently defined.

13. 14.2.135 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    Fill in the blanks in the following table: (i) P(A)=1/3, P(B)=1/5, P(A intersection B)=1/15, P(A union B)=? (ii) P(A)=0.35, P(B)=?, P(A intersection B)=0.25, P(A union B)=0.6 (iii) P(A)=0.5, P(B)=0.35, P(A intersection B)=?, P(A union B)=0.7.

    Hint. Use the addition rule P(A union B)=P(A)+P(B)-P(A intersection B) and rearrange for whichever quantity is missing.

    (i) P(A union B) = 1/3+1/5-1/15 = 5/15+3/15-1/15 = 7/15. (ii) P(B) = P(A union B)-P(A)+P(A intersection B) = 0.6-0.35+0.25 = 0.5. (iii) P(A intersection B) = P(A)+P(B)-P(A union B) = 0.5+0.35-0.7 = 0.15.

    ✦ Working through each part gives: (i) P(A union B)=7/15 (ii) P(B)=0.5 (iii) P(A intersection B)=0.15.

14. 14.2.142 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    Given P(A) = 3/5 and P(B) = 1/5. Find P(A or B), if A and B are mutually exclusive events.

    Hint. For mutually exclusive events, the addition rule simplifies to a direct sum.

    Since A and B are mutually exclusive, P(A intersection B)=0, so P(A or B)=P(A)+P(B)=3/5+1/5.

    ✦ Working through each part gives: P(A or B) = 4/5.

15. 14.2.155 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    If E and F are events such that P(E)=1/4, P(F)=1/2, and P(E and F)=1/8, find (i) P(E or F) (ii) P(not E and not F).

    Hint. Apply the addition rule for part (i); for part (ii), recognise that 'not E and not F' is the complement of 'E or F' by De Morgan's law.

    (i) P(E or F) = P(E)+P(F)-P(E and F) = 1/4+1/2-1/8 = 2/8+4/8-1/8 = 5/8. (ii) 'not E and not F' = (E or F)', so P(not E and not F) = 1 - P(E or F) = 1 - 5/8 = 3/8.

    ✦ Working through each part gives: (i) P(E or F) = 5/8 (ii) P(not E and not F) = 3/8.

16. 14.2.164 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.

    Hint. 'Not E or not F' is E' union F', which by De Morgan's law equals (E intersection F)'.

    P(not E or not F) = P((E intersection F)') = 1 - P(E intersection F) = 0.25, so P(E intersection F) = 0.75, which is not zero.

    ✦ Working through each part gives: E and F are not mutually exclusive, since P(E and F) = 0.75, not 0.

17. 14.2.174 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    A and B are events such that P(A)=0.42, P(B)=0.48, and P(A and B)=0.16. Determine (i) P(not A) (ii) P(not B) (iii) P(A or B).

    Hint. Apply the complement rule for (i) and (ii), and the addition rule for (iii).

    (i) P(not A) = 1-0.42 = 0.58. (ii) P(not B) = 1-0.48 = 0.52. (iii) P(A or B) = 0.42+0.48-0.16 = 0.74.

    ✦ Working through each part gives: (i) 0.58 (ii) 0.52 (iii) 0.74.

18. 14.2.183 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    In Class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

    Hint. Apply the addition rule directly with the given percentages as probabilities.

    P(Maths)=0.4, P(Biology)=0.3, P(both)=0.1. P(Maths or Biology) = 0.4+0.3-0.1.

    ✦ Working through each part gives: P(Mathematics or Biology) = 0.6.

19. 14.2.194 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?

    Hint. Use the addition rule, treating 'at least one' as P(A or B), and solve for the intersection.

    P(A or B) = P(A)+P(B)-P(A and B), so 0.95 = 0.8+0.7-P(A and B), giving P(A and B) = 1.5-0.95.

    ✦ Working through each part gives: P(passing both) = 0.55.

20. 14.2.204 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?

    Hint. Use the complement of 'passing neither' to get P(at least one), then apply the addition rule.

    P(passing neither) = 0.1, so P(passing at least one) = 1-0.1 = 0.9. By the addition rule: 0.9 = P(English)+P(Hindi)-P(both) = 0.75+P(Hindi)-0.5.

    ✦ Working through each part gives: P(passing Hindi) = 0.9-0.75+0.5 = 0.65.

21. 14.2.216 marksNCERT Class 11 Mathematics, Probability, Reprint 2026-27

    In a class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that (i) the student opted for NCC or NSS (ii) the student has opted neither NCC nor NSS (iii) the student has opted NSS but not NCC.

    Hint. Let A='opted NCC', B='opted NSS'; use the addition rule for (i), the complement for (ii), and B-A for (iii).

    P(A)=30/60, P(B)=32/60, P(A and B)=24/60. (i) P(A or B) = 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(B)-P(A and B) = 32/60-24/60 = 8/60 = 2/15.

    ✦ Working through each part gives: (i) 19/30 (ii) 11/30 (iii) 2/15.

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.