CBSE Class 12 Mathematics Probability MCQs Set 10

Answer: (b) 0.44

Explanation:
1. P(X fails) = 0.3, P(Y fails) = 0.2; the two results are independent.
2. P(both fail) = 0.3 × 0.2 = 0.06
3. P(X or Y fails) = 0.3 + 0.2 − 0.06 = 0.44

Teacher's Note:
1. Do not simply add 0.3 and 0.2; the overlap where both fail is counted twice and must be subtracted.
2. A quick check is 1 − P(neither fails) = 1 − 0.7 × 0.8 = 0.44.

Answer: (a) \( P(A) + P(B) - P(A)P(B) \)

Explanation:
1. Addition theorem: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \) (here A + B means A ∪ B).
2. For independent events \( P(A \cap B) = P(A)P(B) \).
3. Hence \( P(A + B) = P(A) + P(B) - P(A)P(B) \).

Teacher's Note:
1. In older notation A + B stands for A ∪ B and AB stands for A ∩ B.
2. The formula can also be written as 1 − P(A′)P(B′), which is often faster for independent events.

Answer: (b) Mutually exclusive

Explanation:
1. \( P(A) = 1 - P(A^c) = 1 - \frac{5}{6} = \frac{1}{6} \)
2. \( P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{6} + \frac{2}{3} - \frac{5}{6} = 0 \), so A and B are mutually exclusive.
3. \( P(A)P(B) = \frac{1}{6} \times \frac{2}{3} = \frac{1}{9} \neq 0 \), so they are not independent; also \( P(A \cup B) \neq 1 \), so not exhaustive.
4. Hence A and B are mutually exclusive (and dependent), which is option (b).

Teacher's Note:
1. Two events with non-zero probabilities can never be both mutually exclusive and independent, because P(A ∩ B) = 0 cannot equal P(A)P(B) > 0.
2. Exhaustive means P(A ∪ B) = 1; always compare the given union with 1 before choosing that option.

Answer: (c) Independent

Explanation:
1. \( P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{2}{3} + \frac{1}{2} - \frac{5}{6} = \frac{1}{3} \)
2. \( P(A)P(B) = \frac{2}{3} \times \frac{1}{2} = \frac{1}{3} \)
3. Since \( P(A \cap B) = P(A)P(B) \), A and B are independent; \( P(A \cup B) = \frac{5}{6} \neq 1 \), so not exhaustive.

Teacher's Note:
1. To test independence, always find P(A ∩ B) from the addition theorem and compare it with P(A)P(B).
2. Since P(A ∩ B) ≠ 0 here, the events cannot be mutually exclusive.

Answer: (d) 29/40

Explanation:
1. P(win toss) = \( \frac{3}{4} \), P(lose toss) = \( \frac{1}{4} \)
2. P(victory) = \( \frac{3}{4} \times \frac{4}{5} + \frac{1}{4} \times \frac{1}{2} \)
3. \( = \frac{3}{5} + \frac{1}{8} = \frac{24 + 5}{40} = \frac{29}{40} \)

Teacher's Note:
1. Split the outcome by the first stage (toss won or lost) and add the two branches.
2. This is the law of total probability in its simplest form.

Answer: (b) 0.54

Explanation:
1. They say the same thing if both speak the truth or both lie.
2. P(both true) = 0.6 × 0.7 = 0.42
3. P(both lie) = 0.4 × 0.3 = 0.12
4. P(same statement) = 0.42 + 0.12 = 0.54

Teacher's Note:
1. Agreement means both true or both false; contradiction means exactly one of them is truthful.
2. As a check, P(contradict) = 0.6 × 0.3 + 0.4 × 0.7 = 0.46, and 0.54 + 0.46 = 1.

Answer: (d) 2/3

Explanation:
1. A can win on throws 1, 3, 5, …: P(A wins) = \( \frac{1}{2} + \left(\frac{1}{2}\right)^2 \frac{1}{2} + \left(\frac{1}{2}\right)^4 \frac{1}{2} + \ldots \)
2. This is a G.P. with first term \( \frac{1}{2} \) and ratio \( \frac{1}{4} \).
3. P(A wins) = \( \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{2}{3} \)

Teacher's Note:
1. For alternate turns with success probability p each turn, the first player wins with probability p / (1 − q²), where q = 1 − p.
2. The player who starts always has the better chance, so 2/3 > 1/2 is a quick sanity check.

Answer: (a) \( \frac{2}{5} \)

Explanation:
1. P(each fails) = \( 1 - \frac{1}{3} = \frac{2}{3} \), \( 1 - \frac{1}{4} = \frac{3}{4} \), \( 1 - \frac{1}{5} = \frac{4}{5} \)
2. P(none solves) = \( \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5} \)

Teacher's Note:
1. For independent people working separately, multiply the failure probabilities to get 'none succeeds'.
2. The probability that the problem is solved would then be 1 − 2/5 = 3/5.

Answer: (a) \( \frac{11}{24} \)

Explanation:
1. Misses: \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \).
2. Only first hits: \( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{6}{24} \); only second: \( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{4} = \frac{3}{24} \); only third: \( \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} = \frac{2}{24} \)
3. P(exactly one hits) = \( \frac{6 + 3 + 2}{24} = \frac{11}{24} \)

Teacher's Note:
1. 'Exactly one' needs three separate cases, each with one hit and two misses.
2. Keep a common denominator (here 24) to add the cases without errors.

Answer: (a) \( \frac{7}{20} \)

Explanation:
1. P(A true) = 0.75, P(B true) = 0.8
2. They contradict when exactly one tells the truth.
3. P = 0.75 × 0.2 + 0.25 × 0.8 = 0.15 + 0.20 = 0.35
4. \( 0.35 = \frac{7}{20} \)

Teacher's Note:
1. Contradiction means one truthful and one lying, so there are exactly two cases to add.
2. P(agree) would be 0.75 × 0.8 + 0.25 × 0.2 = 0.65, and 0.65 + 0.35 = 1 confirms the answer.

Answer: (b) p + q − 2qp

Explanation:
1. Only one alive means exactly one of them dies.
2. P(A dies, B lives) = p(1 − q); P(B dies, A lives) = q(1 − p)
3. Required probability = p − pq + q − pq = p + q − 2pq

Teacher's Note:
1. 'Exactly one' of two independent events always equals P(A) + P(B) − 2P(A)P(B).
2. Do not confuse it with p + q − pq, which is 'at least one' dies.

Answer: (a) 13/32

Explanation:
1. P(white) from boxes: \( \frac{3}{4}, \frac{2}{4}, \frac{1}{4} \); P(black): \( \frac{1}{4}, \frac{2}{4}, \frac{3}{4} \).
2. WWB: \( \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4} = \frac{18}{64} \); WBW: \( \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{6}{64} \); BWW: \( \frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{2}{64} \)
3. Total = \( \frac{26}{64} = \frac{13}{32} \)

Teacher's Note:
1. Decide which box gives the black ball; that gives three mutually exclusive cases.
2. Each case is a product because the draws from different boxes are independent.

Answer: (d) There are infinite values of p, q

Explanation:
1. Success = passes I and (II or III). P(II or III) = \( q + \frac{1}{2} - \frac{q}{2} = \frac{1 + q}{2} \)
2. So P(success) = \( \frac{p(1 + q)}{2} = \frac{1}{2} \), i.e. p(1 + q) = 1.
3. Check options: p = q = 1 gives 2; p = q = \( \frac{1}{2} \) gives \( \frac{3}{4} \); p = 0, q = 1 gives 0. None of these works.
4. For every q in [0, 1], \( p = \frac{1}{1 + q} \) lies in \( \left[\frac{1}{2}, 1\right] \), so there are infinitely many pairs (p, q).

Teacher's Note:
1. When one equation links two unknowns, expect a whole family of solutions rather than a single pair.
2. Always confirm that the values found are valid probabilities, i.e. lie between 0 and 1.

Answer: (d) \( \frac{1}{4} \)

Explanation:
1. By symmetry, every position in the drawing order is equally likely to hold any particular ball.
2. So P(3rd ball red) = P(1st ball red) = \( \frac{2}{8} = \frac{1}{4} \)

Teacher's Note:
1. Without replacement, the unconditional probability for the k-th draw is the same as for the first draw.
2. You can confirm by the long route of adding all cases, but symmetry saves a lot of time in exams.

Answer: (b) \( \frac{25}{56} \)

Explanation:
1. Failures: \( \frac{2}{3}, \frac{5}{7}, \frac{5}{8} \).
2. Only A: \( \frac{1}{3} \cdot \frac{5}{7} \cdot \frac{5}{8} = \frac{25}{168} \); only B: \( \frac{2}{3} \cdot \frac{2}{7} \cdot \frac{5}{8} = \frac{20}{168} \); only C: \( \frac{2}{3} \cdot \frac{5}{7} \cdot \frac{3}{8} = \frac{30}{168} \)
3. Total = \( \frac{75}{168} = \frac{25}{56} \)

Teacher's Note:
1. The common denominator 3 × 7 × 8 = 168 keeps the three cases easy to add.
2. Simplify at the end and match with the options; 75/168 itself is not listed but 25/56 is.

Answer: (d) 76/144

Explanation:
1. Both apples: \( \frac{5}{12} \times \frac{4}{12} = \frac{20}{144} \)
2. Both oranges: \( \frac{7}{12} \times \frac{8}{12} = \frac{56}{144} \)
3. Required = \( \frac{20 + 56}{144} = \frac{76}{144} \)

Teacher's Note:
1. The two cases (both apples, both oranges) are mutually exclusive, so their probabilities are added.
2. Within each case the baskets are independent, so the probabilities are multiplied.

Answer: (c) \( \frac{4}{7} \)

Explanation:
1. First white: \( \frac{4}{7} \cdot \frac{3}{6} \); first black: \( \frac{3}{7} \cdot \frac{4}{6} \)
2. P(2nd white) = \( \frac{12}{42} + \frac{12}{42} = \frac{24}{42} = \frac{4}{7} \)

Teacher's Note:
1. The answer equals P(first ball white), as expected from symmetry of draws without replacement.
2. Use the two-branch method when the question asks only about a later draw.

Answer: (a) \( \leq \frac{1}{4} \)

Explanation:
1. \( \bar{E}_1 \cap \bar{E}_2 \) is the complement of \( E_1 \cup E_2 \) (De Morgan's law).
2. An event and its complement have no common outcome, so \( (E_1 \cup E_2) \cap (\bar{E}_1 \cap \bar{E}_2) = \phi \).
3. Hence the probability is 0, which is \( \leq \frac{1}{4} \).

Teacher's Note:
1. Simplify the set expression with De Morgan's laws before using any probability formula.
2. The probability of an empty event is 0 whatever the values of P(E₁) and P(E₂).

Answer: (a) \( \frac{1}{5} \)

Explanation:
1. Let P(A) = a, P(B) = b. Then a(1 − b) = \( \frac{3}{25} \) and (1 − a)b = \( \frac{8}{25} \).
2. Subtracting: b − a = \( \frac{5}{25} = \frac{1}{5} \), so b = a + \( \frac{1}{5} \).
3. Then \( a - a\left(a + \frac{1}{5}\right) = \frac{3}{25} \) gives \( 25a^2 - 20a + 3 = 0 \), i.e. (5a − 1)(5a − 3) = 0.
4. So P(A) = \( \frac{1}{5} \) or \( \frac{3}{5} \); only \( \frac{1}{5} \) is among the options.

Teacher's Note:
1. Subtracting the two equations removes the product term ab and gives a linear relation instantly.
2. A quadratic may give two valid probabilities, so check which one appears in the options.

Answer: (d) 0.3

Explanation:
1. P(neither) = \( P(A' \cap B') = P(A')P(B') \) for independent events.
2. = (1 − 0.40)(1 − 0.50) = 0.6 × 0.5 = 0.3

Teacher's Note:
1. Complements of independent events are also independent, so their probabilities can be multiplied.
2. Check: P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7, and 1 − 0.7 = 0.3.

Answer: (a) \( \frac{1}{169} \)

Explanation:
1. With replacement the two draws are independent.
2. P(ace each time) = \( \frac{4}{52} = \frac{1}{13} \)
3. P(two aces) = \( \frac{1}{13} \times \frac{1}{13} = \frac{1}{169} \)

Teacher's Note:
1. 'With replacement' keeps the pack the same, so the second probability does not change.
2. Without replacement the answer would be (4/52)(3/51) = 1/221, a common distractor.

Answer: (a) \( \frac{9}{20} \)

Explanation:
1. Total cards = 16, aces = 4, non-aces = 12.
2. P(no ace) = \( \frac{^{12}C_2}{^{16}C_2} = \frac{66}{120} = \frac{11}{20} \)
3. P(at least one ace) = \( 1 - \frac{11}{20} = \frac{9}{20} \)

Teacher's Note:
1. For 'at least one', the complement 'none' is almost always quicker.
2. Notice the pack here has only 16 cards, not 52; read the sample space carefully.

Answer: (c) \( \frac{4}{663} \)

Explanation:
1. P(1st king) = \( \frac{4}{52} \)
2. P(2nd queen | 1st king) = \( \frac{4}{51} \) (all 4 queens remain).
3. P = \( \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663} \)

Teacher's Note:
1. Removing a king does not reduce the number of queens, so the numerator stays 4 while the total drops to 51.
2. Order matters here: 'first king, second queen' is one specific order.

Answer: (c) 1/221

Explanation:
1. P(1st ace) = \( \frac{4}{52} \); P(2nd ace | 1st ace) = \( \frac{3}{51} \)
2. P(both aces) = \( \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \)
3. By the same working, P(both kings) is also 1/221.

Teacher's Note:
1. Without replacement, both the favourable count and the total decrease by 1 for the second draw.
2. Equivalently, ⁴C₂/⁵²C₂ = 6/1326 = 1/221.

Answer: (a) 3/4

Explanation:
1. P(nobody solves) = \( \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4} \)
2. P(problem solved) = \( 1 - \frac{1}{4} = \frac{3}{4} \)

Teacher's Note:
1. 'Problem is solved' means at least one student solves it, so use 1 − P(none).
2. Adding 1/2 + 1/3 + 1/4 exceeds 1, which shows why simple addition is wrong.

Answer: (b) 1/12

Explanation:
1. P(head) = \( \frac{1}{2} \), P(6) = \( \frac{1}{6} \); coin and die are independent.
2. P(head and 6) = \( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)

Teacher's Note:
1. The combined sample space has 2 × 6 = 12 equally likely outcomes, and only (H, 6) is favourable.
2. Independent experiments always combine by multiplication.

Answer: (c) \( \frac{1}{4} \)

Explanation:
1. Given the first two tosses are tails, the coin goes on to tosses 3, 4, 5.
2. It is tossed 5 times exactly when tosses 3 and 4 are both tails (toss 5 happens whatever its result).
3. P = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

Teacher's Note:
1. Under the condition, the first two tosses are already known, so they contribute nothing further.
2. The fifth toss happens whether it is head or tail, so it does not enter the product.

Answer: (a) \( \frac{1}{969} \)

Explanation:
1. Total balls = 20, white = 5.
2. P(4 white) = \( \frac{5}{20} \cdot \frac{4}{19} \cdot \frac{3}{18} \cdot \frac{2}{17} = \frac{120}{116280} = \frac{1}{969} \)

Teacher's Note:
1. The same answer comes from ⁵C₄/²⁰C₄ = 5/4845 = 1/969.
2. Cancel factors step by step to avoid large multiplications.

Answer: (d) \( \frac{4}{19} \)

Explanation:
1. Even numbers from 1 to 19: 2, 4, …, 18, i.e. 9 tickets.
2. P(both even) = \( \frac{9}{19} \times \frac{8}{18} = \frac{72}{342} = \frac{4}{19} \)

Teacher's Note:
1. Count even numbers carefully: from 1 to 19 there are 9 even and 10 odd numbers.
2. Without replacement, the second fraction is 8/18, not 9/19.

Answer: (d) All of these

Explanation:
1. \( P(A/B) = P(A) = \frac{1}{4} \), so A and B are independent.
2. \( P(A'/B) = 1 - P(A/B) = \frac{3}{4} \)
3. \( P(B) = P(B/A) = \frac{1}{2} \) (independence), so \( P(B'/A') = P(B') = \frac{1}{2} \).
4. All three statements are true, so the answer is (d).

Teacher's Note:
1. If P(A/B) = P(A), the events are independent, and so are their complements.
2. P(A′/B) = 1 − P(A/B) always holds because A and A′ split the reduced sample space B.

Answer: (c) 1/2

Explanation:
1. By definition, \( P(B/A) = \frac{P(A \cap B)}{P(A)} \) (A is the given event).
2. \( P(B/A) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \)

Teacher's Note:
1. P(B/A) divides by P(A), the event that is given, not by P(B).
2. Dividing by P(B) instead would give 3/4, which is P(A/B).

Answer: (c) 1/52

Explanation:
1. With replacement, the draws are independent.
2. P(diamond) = \( \frac{13}{52} = \frac{1}{4} \), P(king) = \( \frac{4}{52} = \frac{1}{13} \)
3. P = \( \frac{1}{4} \times \frac{1}{13} = \frac{1}{52} \)

Teacher's Note:
1. Because the card is replaced, the king of diamonds causes no overlap problem here.
2. Without replacement the working would need two cases depending on whether the first card is the king of diamonds.

Answer: (d) 9/25

Explanation:
1. P(no test in a class) = \( 1 - \frac{1}{5} = \frac{4}{5} \)
2. P(no test in both missed classes) = \( \left(\frac{4}{5}\right)^2 = \frac{16}{25} \)
3. P(misses at least one test) = \( 1 - \frac{16}{25} = \frac{9}{25} \)

Teacher's Note:
1. 'At least one' over repeated independent trials is best found as 1 − (failure)ⁿ.
2. Option 7/5 is greater than 1 and can be rejected at once.

Answer: (d) All of these

Explanation:
1. Independence of E and F implies independence of E and \( F^c \), and of \( E^c \) and \( F^c \).
2. So \( P(E/F) = P(E) \) and \( P(E^c/F^c) = P(E^c) \).
3. \( P(E/F) + P(E^c/F^c) = P(E) + P(E^c) = 1 \)
4. All three statements are true.

Teacher's Note:
1. Once two events are independent, every pair formed from them or their complements is independent too.
2. With independence, a conditional probability simply equals the unconditional one.

Answer: (a) 15/16

Explanation:
1. P(no tail in 4 throws) = \( \left(\frac{1}{2}\right)^4 = \frac{1}{16} \)
2. P(at least one tail) = \( 1 - \frac{1}{16} = \frac{15}{16} \)

Teacher's Note:
1. Only the outcome HHHH has no tail, so the complement is a single outcome out of 16.
2. For n throws of a coin, P(at least one tail) = 1 − 1/2ⁿ.

Answer: (c) 16/625

Explanation:
1. The last digit of the product depends only on the last digits of the four numbers.
2. The product ends in 1, 3, 7 or 9 only if every number ends in 1, 3, 7 or 9 (an even digit or 5 would make it even or a multiple of 5).
3. P(one number ends in 1, 3, 7 or 9) = \( \frac{4}{10} = \frac{2}{5} \)
4. Required = \( \left(\frac{2}{5}\right)^4 = \frac{16}{625} \)

Teacher's Note:
1. Any factor ending in 0, 2, 4, 6 or 8 makes the product even, and any factor ending in 5 makes it end in 0 or 5.
2. Products of numbers ending in 1, 3, 7, 9 always end in 1, 3, 7 or 9, so the condition is exact.

Answer: (a) 0.25

Explanation:
1. P(white from bag 1) = \( \frac{4}{6} = \frac{2}{3} \); P(white from bag 2) = \( \frac{3}{8} \)
2. P(both white) = \( \frac{2}{3} \times \frac{3}{8} = \frac{1}{4} = 0.25 \)

Teacher's Note:
1. Draws from different bags are independent, so multiply.
2. Convert fractions to decimals only at the end to match the options.

Answer: (d) \( 1 - (1 - p)^{16} \)

Explanation:
1. The number is correct only if all 16 bits are correct: probability \( (1 - p)^{16} \).
2. P(incorrect number) = \( 1 - (1 - p)^{16} \)

Teacher's Note:
1. A number is wrong if at least one bit is wrong, so use the complement 'all bits correct'.
2. Option p¹⁶ is the chance that all bits are wrong, a very different event.

Answer: (a) \( \frac{3}{5} \)

Explanation:
1. Treat A's and B's wins as independent events with P(A) = \( \frac{1}{5} \) and P(B) = \( \frac{1}{4} \).
2. P(A does not win) = \( \frac{4}{5} \), P(B does not win) = \( \frac{3}{4} \)
3. P(neither wins) = \( \frac{4}{5} \times \frac{3}{4} = \frac{3}{5} \)

Teacher's Note:
1. For independent events, P(neither) is the product of the two complements.
2. Check: P(at least one wins) = 1/5 + 1/4 − 1/20 = 2/5, and 1 − 2/5 = 3/5.

Answer: (d) None of these

Explanation:
1. Least number is 5 means all three chits are from {5, 6, 7} but not all from {6, 7}.
2. P = \( \left(\frac{3}{7}\right)^3 - \left(\frac{2}{7}\right)^3 = \frac{27 - 8}{343} = \frac{19}{343} \)
3. Options (a), (b), (c) do not equal \( \frac{19}{343} \), so the answer is 'None of these'.

Teacher's Note:
1. 'Least is exactly 5' = 'least is at least 5' minus 'least is at least 6'.
2. This subtraction trick works for any 'maximum' or 'minimum is exactly k' problem with replacement.

Answer: (c) \( \frac{1}{9} \)

Explanation:
1. Given max ≤ 10, both tickets come from 1 to 10: \( ^{10}C_2 = 45 \) equally likely pairs.
2. Minimum is 5: one ticket is 5 and the other is from 6 to 10, i.e. 5 pairs.
3. P = \( \frac{5}{45} = \frac{1}{9} \)

Teacher's Note:
1. A condition reduces the sample space; here it shrinks from 100 tickets to 10.
2. Count pairs, not ordered draws, consistently in both numerator and denominator.

Answer: (b) \( \frac{5}{38} \)

Explanation:
1. P(1st I) = \( \frac{10}{20} \), P(2nd I) = \( \frac{9}{19} \), P(3rd T) = \( \frac{10}{18} \)
2. P(IIT) = \( \frac{10}{20} \cdot \frac{9}{19} \cdot \frac{10}{18} = \frac{900}{6840} = \frac{5}{38} \)

Teacher's Note:
1. After two I-cards are taken, all 10 T-cards remain out of 18, so the third fraction is 10/18.
2. Cancel early: (1/2)(9/19)(5/9) = 5/38.