A bag has 4 white and 6 black balls. Two balls are drawn without replacement. Given the first ball drawn is white, the probability the second is also white is:
(a) 4/10
(b) 3/9
(c) 4/9
(d) 2/5
✓ Correct Answer: (b) 3/9
Solution: After drawing one white ball, 3 white and 6 black balls remain (9 total). P(2nd white | 1st white) = 3/9 = 1/3
Question 10
A bag contains 3 red and 5 black balls. Two balls are drawn one after the other without replacement. The probability that both are red is:
Solution: P(B|A) = P(A∩B)/P(A). This is defined only when the conditioning event A has non-zero probability, i.e. P(A) ≠ 0.
Question 16
A fair die is thrown once. The probability of getting a prime number is:
(a) 1/6
(b) 1/3
(c) 1/2
(d) 2/3
✓ Correct Answer: (c) 1/2
Solution: Prime numbers on a die: {2, 3, 5} → 3 outcomes out of 6. P = 3/6 = 1/2
Question 17
If P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2, then P(A|B) is:
(a) 0.4
(b) 0.5
(c) 0.2
(d) 0.8
✓ Correct Answer: (b) 0.5
Solution: P(A|B) = P(A∩B)/P(B) = 0.2/0.4 = 0.5
Question 18
Two cards are drawn one by one without replacement from a pack of 52. The probability that the second card is a king, given the first was also a king, is:
(a) 4/52
(b) 3/51
(c) 1/13
(d) 1/17
✓ Correct Answer: (b) 3/51
Solution: After drawing one king, 3 kings remain in 51 cards. P(2nd king | 1st king) = 3/51 = 1/17
Question 19
A card is drawn from a pack of 52. What is the probability that it is either a heart or a king?
Solution: If A ⊂ B, then A∩B = A. So P(A|B) = P(A∩B)/P(B) = P(A)/P(B).
Question 21
From a group of 3 boys and 2 girls, two students are selected at random. The probability that at least one is a girl is:
(a) 3/10
(b) 7/10
(c) 1/10
(d) 9/10
✓ Correct Answer: (b) 7/10
Solution: Total ways = ⁵C₂ = 10. Ways with no girl (both boys) = ³C₂ = 3. P(at least one girl) = 1 − 3/10 = 7/10
Question 22
A speaks truth in 60% and B in 80% of cases. The probability that they will contradict each other in stating a fact is:
(a) 0.36
(b) 0.44
(c) 0.48
(d) 0.52
✓ Correct Answer: (b) 0.44
Solution: Contradiction occurs when one tells truth and other lies. P = P(A truth)·P(B lie) + P(A lie)·P(B truth) = (0.6)(0.2) + (0.4)(0.8) = 0.12 + 0.32 = 0.44
Question 23
For two events A and B, P(A) = 1/2, P(B) = 1/3. If A and B are mutually exclusive, then P(A∪B) is:
An urn contains 6 red, 4 blue, and 2 green balls. A ball is drawn at random. The probability of drawing a non-red ball is:
(a) 1/2
(b) 1/3
(c) 2/3
(d) 3/4
✓ Correct Answer: (a) 1/2
Solution: Total = 12 balls. Red = 6. Non-red = 6. P(non-red) = 6/12 = 1/2
Question 25
If A and B are independent events with P(A) = 0.3 and P(B) = 0.4, then P(A' ∩ B') is:
(a) 0.12
(b) 0.42
(c) 0.58
(d) 0.70
✓ Correct Answer: (b) 0.42
Solution: Since A and B are independent, A' and B' are also independent. P(A') = 0.7, P(B') = 0.6 P(A'∩B') = P(A') × P(B') = 0.7 × 0.6 = 0.42
Question 26
If P(A) = 0.6 and P(A∩B) = 0.3, then P(B|A) is:
(a) 0.18
(b) 0.3
(c) 0.5
(d) 2
✓ Correct Answer: (c) 0.5
Solution: P(B|A) = P(A∩B)/P(A) = 0.3/0.6 = 0.5
Question 27
A die is thrown twice. Given that the sum of outcomes is 8, the probability that both outcomes were 4 is:
(a) 1/5
(b) 1/6
(c) 1/36
(d) 5/36
✓ Correct Answer: (a) 1/5
Solution: Ways to get sum 8: {(2,6),(3,5),(4,4),(5,3),(6,2)} → 5 outcomes. Favourable (both 4): {(4,4)} → 1 outcome. P(both 4 | sum = 8) = 1/5
📋 Assertion-Reason Questions
Statement I is the Assertion (A) and Statement II is the Reason (R). Choose:
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
Assertion-Reason 1
Assertion (A): If A and B are mutually exclusive events, they cannot be independent. Reason (R): For independent events, P(A∩B) = P(A)·P(B), while for mutually exclusive events P(A∩B) = 0.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Solution: If A and B are mutually exclusive (P(A∩B) = 0), they can be independent only if P(A) = 0 or P(B) = 0. So for non-trivial events, they cannot be both mutually exclusive and independent. R correctly explains A.
Assertion-Reason 2
Assertion (A): The probability of the sample space S is always 1. Reason (R): The sample space S is a sure event, meaning it contains all possible outcomes of the experiment.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Solution: P(S) = 1 because the sample space represents the event that some outcome occurs, which is certain. R correctly explains why A is true.
Assertion-Reason 3
Assertion (A): If P(A|B) = P(A), then A and B are dependent events. Reason (R): When occurrence of B does not affect probability of A, the events are independent.
(a) Both A and R are True and R is the correct explanation of A
(b) Both A and R are True but R is not the correct explanation of A
(c) A is True but R is False
(d) A is False but R is True
✓ Correct Answer: (d) A is False but R is True
Solution: If P(A|B) = P(A), it means B's occurrence doesn't affect A → they are INDEPENDENT, not dependent. A is false. R is correctly stated.
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A box contains 5 red and 3 blue balls. Two balls are drawn one at a time without replacement. Find: (i) P(both red), (ii) P(both blue), (iii) P(first red and second blue), (iv) P(second ball is blue).
A coin is tossed three times. Events are defined as: A = at least two heads, B = at most two heads, C = first toss is head. Find: (i) P(A|B), (ii) P(B|C), (iii) Are A and C independent?
Complete Solution:
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}; n(S) = 8 A = {HHH, HHT, HTH, THH}: 4 outcomes; B = {HHT, HTH, THH, HTT, THT, TTH, TTT}: 7 outcomes; C = {HHH, HHT, HTH, HTT}: 4 outcomes
(iii) A∩C = {HHH, HHT, HTH}: 3 outcomes; P(A∩C) = 3/8 P(A)·P(C) = (4/8)(4/8) = 16/64 = 1/4 = 2/8 Since 3/8 ≠ 2/8, A and C are NOT independent.
(i) P(A|B) = 3/7; (ii) P(B|C) = 3/4; (iii) A and C are dependent.
Question 5
Three students A, B, C independently try to solve a problem. Their probabilities of solving it are 1/3, 1/4, and 1/5 respectively. Find the probability that: (i) all three solve it, (ii) none of them solves it, (iii) at least one of them solves it, (iv) exactly two of them solve it.
A problem in mathematics is given to 3 students A, B, C whose chances of solving it are 1/2, 1/3 and 1/4 respectively. Find the probability that (i) the problem is solved, (ii) exactly one solves it.
In an examination, 30% students failed in Physics, 25% in Mathematics, and 10% in both. A student is selected at random. Find the probability that: (i) she failed in Physics or Maths, (ii) she failed in Physics given she failed in Maths.
Complete Solution:
P(P) = 0.30, P(M) = 0.25, P(P∩M) = 0.10
(i) P(P∪M) = 0.30 + 0.25 − 0.10 = 0.45
(ii) P(P|M) = P(P∩M)/P(M) = 0.10/0.25 = 2/5 = 0.4
(i) P(P or M) = 0.45; (ii) P(P | M) = 0.4
Question 9
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both. If one of these students is selected at random, find the probability that: (i) the student opted for NCC or NSS, (ii) the student opted for NSS given they opted for NCC, (iii) the student opted for NCC given they opted for NSS, (iv) the student opted for exactly one of NCC or NSS.
A and B are two independent events. The probability that A occurs is 1/4 and the probability that B occurs is 2/5. Find: (i) P(A∩B), (ii) P(A∪B), (iii) P(A|B), (iv) P(A'∩B'), (v) P(A'∪B').
Complete Solution:
P(A) = 1/4, P(B) = 2/5, P(A') = 3/4, P(B') = 3/5
(i) Since A, B independent: P(A∩B) = P(A)·P(B) = (1/4)(2/5) = 2/20 = 1/10
(iv) A' and B' are also independent: P(A'∩B') = P(A')·P(B') = (3/4)(3/5) = 9/20
(v) P(A'∪B') = 1 − P(A∩B) = 1 − 1/10 = 9/10
(i) 1/10; (ii) 11/20; (iii) 1/4; (iv) 9/20; (v) 9/10
📊
Case Studies
Real-world application based questions
Case Study 1
Card Drawing at a School Fair
At a school fair, a student picks one card at random from a well-shuffled deck of 52 playing cards. The event A is defined as "the card is a red card" and event B is defined as "the card is a face card" (Jack, Queen or King).
(i) What is P(A) and P(B)?
(a) P(A) = 1/4, P(B) = 3/13
(b) P(A) = 1/2, P(B) = 3/13
(c) P(A) = 1/2, P(B) = 1/4
(d) P(A) = 1/4, P(B) = 1/4
✓ Correct Answer: (b) P(A) = 1/2, P(B) = 3/13
Solution: Red cards = 26, so P(A) = 26/52 = 1/2. Face cards = 12 (J, Q, K × 4 suits), so P(B) = 12/52 = 3/13.
(ii) Given the card drawn is a face card, what is the probability it is red?
(iii)(a) Are events A and B independent? Show working.
✓ Yes, A and B are independent
Solution: P(A∩B) = 6/52 = 3/26 P(A)·P(B) = (1/2)(3/13) = 3/26 Since P(A∩B) = P(A)·P(B), events A and B are independent. Knowing the card is a face card does not change the probability that it is red.
(iii)(b) What is P(A∪B) — the probability of drawing a red card or a face card?
A coaching institute has students preparing for two competitive exams — Exam X and Exam Y. The probability that a randomly selected student clears Exam X is 0.5 and Exam Y is 0.4. The probability of clearing both exams is 0.2. A student is selected at random.
(i) What is the probability that the student clears at least one exam?
(ii) Given that the student cleared Exam X, what is the probability they also cleared Exam Y?
(a) 0.2
(b) 0.4
(c) 0.5
(d) 0.8
✓ Correct Answer: (b) 0.4
Solution: P(Y|X) = P(X∩Y)/P(X) = 0.2/0.5 = 0.4
(iii)(a) Are events X and Y independent? Justify.
✓ Yes, X and Y ARE independent
Solution: For independence, we check if P(X∩Y) = P(X)·P(Y). P(X)·P(Y) = 0.5 × 0.4 = 0.20, and P(X∩Y) = 0.20 (given). Since P(X∩Y) = P(X)·P(Y), events X and Y are independent. Clearing one exam does not affect the probability of clearing the other.
(iii)(b) What is the probability that the student cleared exactly one exam?
A bag contains 5 red, 4 white and 3 blue balls. Two balls are drawn one after the other without replacement. Use the multiplication theorem of conditional probability to answer the following questions.
(i) What is the probability that the first ball is red and the second is white?
In a school of 200 students, 120 are from Class XI and 80 are from Class XII. Among the Class XI students, 45 are in the Science stream and 75 are in the Commerce stream. Among Class XII students, 30 are in Science and 50 are in Commerce. A student is selected at random for a school council position.
(i) What is the probability that a randomly selected student is from Class XI?
(a) 2/5
(b) 3/5
(c) 1/2
(d) 7/10
✓ Correct Answer: (b) 3/5
Solution: P(Class XI) = 120/200 = 3/5
(ii) Given that the selected student is in the Science stream, what is the probability they are from Class XI?
(a) 45/75
(b) 3/5
(c) 45/200
(d) 3/4
✓ Correct Answer: (b) 3/5
Solution: Total Science students = 45 + 30 = 75 P(Class XI | Science) = 45/75 = 3/5
(iii)(a) Given that the selected student is from Class XI, what is the probability they are in the Commerce stream?
(a) 75/200
(b) 5/8
(c) 3/8
(d) 3/4
✓ Correct Answer: (b) 5/8
Solution: P(Commerce | Class XI) = 75/120 = 5/8
(iii)(b) Are the events "student is from Class XI" and "student is in Science stream" independent? Justify.
✓ Yes, the two events are independent
Solution: P(XI) = 120/200 = 3/5; P(Science) = 75/200 = 3/8 P(XI ∩ Science) = 45/200 = 9/40 P(XI) × P(Science) = (3/5)(3/8) = 9/40 Since P(XI ∩ Science) = P(XI) × P(Science), the events are independent. A student's class does not influence their likelihood of being in the Science stream in this school.
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