CBSE Class 12 Mathematics Probability MCQs Set 11

Answer: (d) \( \frac{1}{2} \)

Explanation:
1. Primes in A = {2, …, 20}: 2, 3, 5, 7, 11, 13, 17, 19, i.e. 8 primes.
2. Primes greater than 10: 11, 13, 17, 19, i.e. 4.
3. P = \( \frac{4}{8} = \frac{1}{2} \)

Teacher's Note:
1. Since the number is known to be prime, the sample space is the 8 primes, not the 19 numbers.
2. Remember 2 is a prime; leaving it out is a common slip.

Answer: (d) \( 1 - \left(\frac{43}{45}\right)^3 \)

Explanation:
1. Two-digit numbers with digit product 12: 26, 62, 34, 43, i.e. 4 out of 90.
2. P(laugh on one choice) = \( \frac{4}{90} = \frac{2}{45} \), so P(no laugh) = \( \frac{43}{45} \).
3. P(laughs at least once in 3) = \( 1 - \left(\frac{43}{45}\right)^3 \)

Teacher's Note:
1. List digit pairs with product 12 systematically: 2 × 6 and 3 × 4, each in two orders.
2. There are 90 numbers from 10 to 99, not 89 or 100.

Answer: (a) \( \frac{7}{8} \)

Explanation:
1. P(M | W) = 0.7, so \( P(M \cap W) = 0.7 \times 0.5 = 0.35 \)
2. \( P(W | M) = \frac{P(M \cap W)}{P(M)} = \frac{0.35}{0.4} = \frac{7}{8} \)

Teacher's Note:
1. First recover the joint probability from the given conditional, then divide by the new condition.
2. This 'reversal' of a conditional is exactly the idea behind Bayes' theorem.

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

Explanation:
1. P(sum 5) = \( \frac{4}{36} \), P(sum 7) = \( \frac{6}{36} \).
2. Throws that give neither just repeat the game, so only the first throw with 5 or 7 matters.
3. P(5 before 7) = \( \frac{4}{4 + 6} = \frac{2}{5} \)

Teacher's Note:
1. When repeating until A or B occurs, P(A first) = P(A)/(P(A) + P(B)).
2. The same answer comes from the geometric series (4/36)[1 + (26/36) + (26/36)² + …].

Answer: (b) \( \frac{21}{40} \)

Explanation:
1. Red from 1st, black from 2nd: \( \frac{3}{8} \cdot \frac{4}{10} = \frac{12}{80} \)
2. Black from 1st, red from 2nd: \( \frac{5}{8} \cdot \frac{6}{10} = \frac{30}{80} \)
3. Total = \( \frac{42}{80} = \frac{21}{40} \)

Teacher's Note:
1. 'One red and one black' has two orders here because the balls come from different bags.
2. Option 3/20 is only the first case; forgetting the second case is a common error.

Answer: (b) \( \frac{8}{17} \)

Explanation:
1. Sum 9 with two dice: (3,6), (4,5), (5,4), (6,3), so p = \( \frac{4}{36} = \frac{1}{9} \), q = \( \frac{8}{9} \).
2. B wins on throws 2, 4, 6, …: P = \( qp + q^3p + \ldots = \frac{qp}{1 - q^2} \)
3. \( = \frac{\frac{8}{9} \cdot \frac{1}{9}}{1 - \frac{64}{81}} = \frac{\frac{8}{81}}{\frac{17}{81}} = \frac{8}{17} \)

Teacher's Note:
1. The second player wins with probability q/(1 + q), here (8/9)/(17/9) = 8/17.
2. Check: P(A wins) = 9/17 and 9/17 + 8/17 = 1.

Answer: (b) \( \frac{1}{37} \)

Explanation:
1. Let K = knows, G = guesses: P(K) = 0.9, P(G) = 0.1; P(correct | K) = 1, P(correct | G) = \( \frac{1}{4} \).
2. P(correct) = 0.9 × 1 + 0.1 × \( \frac{1}{4} \) = 0.9 + 0.025 = 0.925
3. P(G | correct) = \( \frac{0.025}{0.925} = \frac{25}{925} = \frac{1}{37} \)

Teacher's Note:
1. In Bayes' problems, list each cause with its prior and the chance it produces the observed result.
2. The posterior is one branch divided by the sum of all branches.

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

Explanation:
1. Each urn is chosen with probability \( \frac{1}{3} \); P(red) from urns: \( \frac{6}{10}, \frac{4}{10}, \frac{5}{10} \).
2. P(urn 1 | red) = \( \frac{\frac{1}{3} \cdot \frac{6}{10}}{\frac{1}{3}\left(\frac{6}{10} + \frac{4}{10} + \frac{5}{10}\right)} = \frac{6}{15} = \frac{2}{5} \)

Teacher's Note:
1. When all causes have equal priors, the priors cancel and only the likelihoods matter.
2. Here simply compare red counts 6 : 4 : 5, giving 6/15 for the first urn.

Answer: (b) 45/61

Explanation:
1. P(first group) = \( \frac{3}{5} \), P(second group) = \( \frac{2}{5} \); P(white) = \( \frac{5}{8} \) and \( \frac{2}{6} = \frac{1}{3} \).
2. Branches: \( \frac{3}{5} \cdot \frac{5}{8} = \frac{3}{8} \) and \( \frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15} \)
3. P(first group | white) = \( \frac{\frac{3}{8}}{\frac{3}{8} + \frac{2}{15}} = \frac{45}{45 + 16} = \frac{45}{61} \)

Teacher's Note:
1. The prior for each group is the number of bags in it divided by the total bags (3 out of 5).
2. Multiply by 120 (LCM of 8 and 15) to clear fractions before dividing.

Answer: (c) \( \frac{11}{50} \)

Explanation:
1. Prior: P(lost card heart) = \( \frac{1}{4} \), P(not heart) = \( \frac{3}{4} \).
2. P(two hearts | lost heart) = \( \frac{^{12}C_2}{^{51}C_2} \); P(two hearts | lost non-heart) = \( \frac{^{13}C_2}{^{51}C_2} \)
3. P(lost heart | two hearts) = \( \frac{\frac{1}{4} \cdot 66}{\frac{1}{4} \cdot 66 + \frac{3}{4} \cdot 78} = \frac{66}{300} = \frac{11}{50} \)

Teacher's Note:
1. The common denominator ⁵¹C₂ cancels, so only the counts 66 and 78 need to be compared.
2. Seeing two hearts makes a missing heart less likely, so the answer must be below 1/4, which 11/50 is.

Answer: (b) \( \frac{38}{63} \)

Explanation:
1. If a white ball is moved (prob \( \frac{4}{7} \)), bag 2 has 4 white, 5 black: P(black) = \( \frac{5}{9} \).
2. If a black ball is moved (prob \( \frac{3}{7} \)), bag 2 has 3 white, 6 black: P(black) = \( \frac{6}{9} \).
3. P(black) = \( \frac{4}{7} \cdot \frac{5}{9} + \frac{3}{7} \cdot \frac{6}{9} = \frac{20 + 18}{63} = \frac{38}{63} \)

Teacher's Note:
1. Transfer problems are total-probability problems: split by what was transferred.
2. After the transfer bag 2 has 9 balls, not 8; forgetting this is the usual mistake.

Answer: (c) \( \frac{5}{8} \)

Explanation:
1. The home is unlocked with probability 0.4.
2. If locked (0.6), he gets in only if the right key is among his 3 out of 8: \( \frac{^7C_2}{^8C_3} = \frac{21}{56} = \frac{3}{8} \).
3. P(gets in) = 0.4 + 0.6 × \( \frac{3}{8} \) = 0.4 + 0.225 = 0.625 = \( \frac{5}{8} \)

Teacher's Note:
1. Choosing 3 keys out of 8 includes a particular key with probability 3/8.
2. Do not forget the unlocked case, which lets him in without any key.

Answer: (b) \( \frac{7}{15} \)

Explanation:
1. Each bag is chosen with probability \( \frac{1}{2} \); P(white) = \( \frac{3}{5} \) from x and \( \frac{2}{6} = \frac{1}{3} \) from y.
2. P(white) = \( \frac{1}{2} \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{1}{3} = \frac{3}{10} + \frac{1}{6} = \frac{14}{30} = \frac{7}{15} \)

Teacher's Note:
1. Pooling all balls (5 white of 11) is wrong because the two bags have different sizes.
2. Total probability weights each bag by the chance of choosing it.

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

Explanation:
1. Second win at the third test means exactly one win in the first two tests and a win in the third.
2. P(exactly one win in 2) = \( 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} \)
3. P = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)

Teacher's Note:
1. 'The k-th success occurs on trial n' means k − 1 successes in the first n − 1 trials and a success on trial n.
2. Matches 4 and 5 are irrelevant once the second win has occurred at the third test.

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

Explanation:
1. First red: \( \frac{2}{5} \cdot \frac{1}{4} \); first white: \( \frac{3}{5} \cdot \frac{2}{4} \)
2. P(2nd red) = \( \frac{2}{20} + \frac{6}{20} = \frac{8}{20} = \frac{2}{5} \)

Teacher's Note:
1. The answer equals P(first ball red) = 2/5, as symmetry predicts.
2. Setting up both branches (first red, first white) avoids picking a distractor like 8/25 from a single product.

Answer: (a) \( \frac{33}{48} \)

Explanation:
1. P(nobody solves) = \( \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} = \frac{15}{48} \)
2. P(solved) = \( 1 - \frac{15}{48} = \frac{33}{48} \)

Teacher's Note:
1. 'Question is solved' means at least one student solves it.
2. Keep 48 as the denominator because all options are written over 48.

Answer: (c) \( \frac{13}{32} \)

Explanation:
1. P(girl) from groups: \( \frac{3}{4}, \frac{2}{4}, \frac{1}{4} \); P(boy): \( \frac{1}{4}, \frac{2}{4}, \frac{3}{4} \).
2. GBB: \( \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4} = \frac{18}{64} \); BGB: \( \frac{1}{4} \cdot \frac{2}{4} \cdot \frac{3}{4} = \frac{6}{64} \); BBG: \( \frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{2}{64} \)
3. Total = \( \frac{26}{64} = \frac{13}{32} \)

Teacher's Note:
1. Decide which group gives the girl; that creates three exclusive cases.
2. This has the same structure as the white and black balls from three boxes.

Answer: (b) 2/5

Explanation:
1. P(number > 4) = \( \frac{2}{6} = \frac{1}{3} \) = p, q = \( \frac{2}{3} \).
2. Stopping at toss k has probability \( q^{k-1}p \). Even k: \( qp + q^3p + q^5p + \ldots = \frac{qp}{1 - q^2} \)
3. \( = \frac{\frac{2}{3} \cdot \frac{1}{3}}{1 - \frac{4}{9}} = \frac{\frac{2}{9}}{\frac{5}{9}} = \frac{2}{5} \)

Teacher's Note:
1. Sum only the even-numbered terms of the geometric distribution; the ratio becomes q².
2. Equivalently q/(1 + q) = (2/3)/(5/3) = 2/5.

Answer: (a) \( \frac{3}{4} \)

Explanation:
1. P(head) = \( \frac{1}{2} \); P(5 or 6) = \( \frac{1}{3} \). Sequence: coin, die, coin, die, …
2. One full round with no success has probability \( \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \).
3. P(head first) = \( \frac{1}{2}\left(1 + \frac{1}{3} + \frac{1}{9} + \ldots\right) = \frac{\frac{1}{2}}{1 - \frac{1}{3}} = \frac{3}{4} \)

Teacher's Note:
1. For alternating trials, P(first wins) = p₁/(1 − q₁q₂), where q₁q₂ is the chance a whole round fails.
2. Starting first gives the coin an advantage, so the answer should exceed 1/2.

Answer: (d) 0.6976

Explanation:
1. The plane escapes only if all four shots miss: 0.6 × 0.7 × 0.8 × 0.9 = 0.3024
2. P(plane is hit) = 1 − 0.3024 = 0.6976

Teacher's Note:
1. 'Hits the plane' means at least one of the four shots hits, so use the complement.
2. Adding 0.4 + 0.3 + 0.2 + 0.1 = 1 would wrongly suggest a certain hit.

Answer: (c) 0.95

Explanation:
1. \( P(A \cap B) = P(B)P(A/B) = 0.05p \); \( P(A' \cap B') = 0.95p \), so \( P(A \cap B') = 0.95(1 - p) \).
2. \( P(A) = 0.05p + 0.95(1 - p) = 0.95 - 0.9p \)
3. \( P(B/A) = \frac{0.05p}{0.95 - 0.9p} = 0.5 \) gives 0.05p = 0.475 − 0.45p, so 0.5p = 0.475.
4. p = 0.95

Teacher's Note:
1. Break P(A) into the parts inside B and inside B′; each comes from a given conditional probability.
2. Substituting p = 0.95 back gives P(A) = 0.095 and P(B/A) = 0.0475/0.095 = 0.5, a useful check.

Answer: (a) Independent

Explanation:
1. P(A) = P(B) = P(C) = \( \frac{4}{8} = \frac{1}{2} \)
2. Each pair shares 2 tickets, e.g. A ∩ B = {000, 001}: \( P(A \cap B) = \frac{1}{4} = P(A)P(B) \); similarly for the other pairs.
3. \( A \cap B \cap C \) = {000}: \( P = \frac{1}{8} = P(A)P(B)P(C) \)
4. So A, B, C are (mutually) independent; they are not mutually exclusive since 000 lies in all three.

Teacher's Note:
1. Mutual independence of three events needs all three pairwise conditions and the triple product condition.
2. Events with positive probability that are independent can never be mutually exclusive.

Answer: (d) All of these

Explanation:
1. Strictly increasing triples: \( ^6C_3 = 20 \), so \( P(E_1) = \frac{20}{216} \).
2. Increasing A.P.s: 123, 234, 345, 456, 135, 246, i.e. 6, so \( P(E_2) = \frac{6}{216} = \frac{1}{36} \).
3. \( E_2 \subset E_1 \), so \( P(E_2) \leq P(E_1) \) and \( P(E_2 \cap E_1) = P(E_2) = \frac{1}{36} \).
4. \( P(E_2 | E_1) = \frac{6}{20} = \frac{3}{10} \); all of (a), (b), (c) hold.

Teacher's Note:
1. Any three distinct numbers can be arranged in increasing order in exactly one way, so strictly increasing outcomes number ⁶C₃.
2. When one event is contained in another, their intersection is the smaller event; also here P(E₁) = (10/3)P(E₂).

Answer: (b) 0.19

Explanation:
1. Let \( p_i \) be the given probabilities. Required = P(exactly 6) + P(exactly 7) + P(all 8).
2. P(all 8) = 0.2 × 0.3 × … × 0.9 = 0.0036 (approx.)
3. Adding the cases with one failure gives P(exactly 7) ≈ 0.0373, and the cases with two failures give P(exactly 6) ≈ 0.1460.
4. Total ≈ 0.0036 + 0.0373 + 0.1460 = 0.1869 ≈ 0.19

Teacher's Note:
1. When probabilities differ, 'at least six of eight' must be built from separate exactly-k cases.
2. A useful shortcut: P(exactly 7) = P(all 8) × Σ(1 − pᵢ)/pᵢ, since each term replaces one success by a failure.

Answer: (a) 5/21

Explanation:
1. Given face 1 or 2: P(1 or 2) = 0.1 + 0.32 = 0.42
2. P(face 1 | 1 or 2) = \( \frac{0.1}{0.42} = \frac{10}{42} = \frac{5}{21} \)

Teacher's Note:
1. The condition reduces the sample space to faces 1 and 2 only.
2. For a biased die, use the given probabilities, never counting of faces.

Answer: (c) 0.75

Explanation:
1. P(even) = P(2) + P(4) + P(6) = 0.24 + 0.18 + 0.14 = 0.56
2. P(2 or 4) = 0.24 + 0.18 = 0.42
3. P(2 or 4 | even) = \( \frac{0.42}{0.56} = 0.75 \)

Teacher's Note:
1. Faces 2 and 4 are already even, so the intersection with 'even' is just {2, 4}.
2. Check: P(6 | even) = 0.14/0.56 = 0.25 and 0.75 + 0.25 = 1.

Answer: (c) 8/15

Explanation:
1. P(white) = \( \frac{1}{2} \cdot \frac{2}{5} + \frac{1}{2} \cdot \frac{4}{6} \)
2. \( = \frac{1}{5} + \frac{1}{3} = \frac{8}{15} \)

Teacher's Note:
1. Weight each bag by 1/2 because one bag is chosen at random.
2. Pooling all balls (6 white of 11) is incorrect since the bags have different sizes.

Answer: (a) 9/16

Explanation:
1. P(spade) = \( \frac{13}{52} = \frac{1}{4} \), P(not spade) = \( \frac{3}{4} \); draws are independent (replacement).
2. P(fails first two times) = \( \frac{3}{4} \times \frac{3}{4} = \frac{9}{16} \)

Teacher's Note:
1. Only the first two draws matter for 'fails the first two times'.
2. If the question asked for the first spade exactly on the third draw, the answer would be (9/16)(1/4) = 9/64.

Answer: (a) \( P\left(\frac{A}{B}\right) \geq \frac{P(A) + P(B) - 1}{P(B)},\ P(B) \neq 0 \) is always true

Explanation:
1. \( P(A \cap B) = P(A) + P(B) - P(A \cup B) \geq P(A) + P(B) - 1 \), since \( P(A \cup B) \leq 1 \).
2. Dividing by P(B) > 0: \( P(A/B) \geq \frac{P(A) + P(B) - 1}{P(B)} \), so (a) is always true.
3. (b) is false because \( P(A \cap \bar{B}) = P(A) - P(A \cap B) \) always holds; (c) is false because for disjoint events \( P(A \cup B) = P(A) + P(B) \), not \( 1 - P(\bar{A})P(\bar{B}) \).

Teacher's Note:
1. The bound P(A ∩ B) ≥ P(A) + P(B) − 1 (Bonferroni inequality) is useful whenever only P(A) and P(B) are known.
2. 1 − P(Ā)P(B̄) is the union formula for independent events, not for disjoint events.

Answer: (d) 0.63

Explanation:
1. P(new product) = P(A)P(N|A) + P(B)P(N|B) + P(C)P(N|C)
2. = 0.5 × 0.7 + 0.3 × 0.6 + 0.2 × 0.5
3. = 0.35 + 0.18 + 0.10 = 0.63

Teacher's Note:
1. The three groups are mutually exclusive and exhaustive, so total probability applies directly.
2. The individual terms 0.35, 0.18 and 0.10 appear as distractors; the answer is their sum.

Answer: (a) \( P(E/F) + P(\bar{E}/F) = 1 \)

Explanation:
1. Given F, either E or \( \bar{E} \) occurs, so \( P(E/F) + P(\bar{E}/F) = \frac{P(E \cap F) + P(\bar{E} \cap F)}{P(F)} = \frac{P(F)}{P(F)} = 1 \); (a) is true.
2. (b) and (c) mix two different conditions F and \( \bar{F} \), so they fail in general; e.g. if E is the sure event, (b) gives 1 + 1 = 2, and if E = F, (c) gives 0 + 0 = 0.
3. (d): if E = F, \( P(\bar{E}/\bar{F}) + P(E/F) = 1 + 1 = 2 \neq 1 \), so (d) is false.
4. Hence only (a) is always true.

Teacher's Note:
1. The complement rule P(E/F) + P(Ē/F) = 1 works only when both terms are conditioned on the same event.
2. To reject a general identity, one counter-example such as E = F is enough.

Answer: (a) Both \( S_1 \) and \( S_2 \) are true

Explanation:
1. Mutual independence gives \( P(A \cap B \cap C) = P(A)P(B)P(C) = P(A)P(B \cap C) \), so \( S_2 \) is true.
2. \( P(A \cap (B \cup C)) = P(A \cap B) + P(A \cap C) - P(A \cap B \cap C) \)
3. \( = P(A)[P(B) + P(C) - P(B)P(C)] = P(A)P(B \cup C) \), so \( S_1 \) is true.
4. Hence both \( S_1 \) and \( S_2 \) are true.

Teacher's Note:
1. If A, B, C are mutually independent, A is independent of any event formed from B and C.
2. Pairwise independence alone is not enough for this; the triple product condition is needed.

Answer: (b) 3/8

Explanation:
1. P(brown hair) = 0.40, P(brown hair and brown eyes) = 0.15
2. P(brown eyes | brown hair) = \( \frac{0.15}{0.40} = \frac{3}{8} \)

Teacher's Note:
1. Divide by the probability of the given condition (brown hair), not by P(brown eyes).
2. Dividing by 0.25 would give 3/5, which is P(brown hair | brown eyes).

Answer: (c) \( P(A \cup B)^c = P(A^c)P(B^c) \)

Explanation:
1. The given condition says \( P(A \cap B) = P(A)P(B) \), so A and B are independent.
2. Then \( A^c \) and \( B^c \) are independent: \( P((A \cup B)^c) = P(A^c \cap B^c) = P(A^c)P(B^c) \), so (c) is true.
3. (a) is false since \( P(B/A) = P(B) \); (b) is false since \( A^c \cap B^c \) has probability \( P(A^c)P(B^c) \gt 0 \).
4. (d) is false since \( P(A \cap B) = P(A)P(B) \gt 0 \).

Teacher's Note:
1. Comparing the given union formula with the addition theorem immediately gives P(A ∩ B) = P(A)P(B).
2. Independent events with probabilities strictly between 0 and 1 always overlap, so P(A ∩ B) > 0.

Answer: (b) \( \frac{5}{36} \)

Explanation:
1. From 12 games: P(A wins) = \( \frac{6}{12} = \frac{1}{2} \), P(B wins) = \( \frac{4}{12} = \frac{1}{3} \).
2. Winning alternately in 3 games: ABA or BAB.
3. P = \( \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36} \)

Teacher's Note:
1. 'Win alternately' allows either player to start, so both patterns must be counted.
2. A draw breaks the alternation, so no draw can occur in these patterns.

Answer: (a) 10

Explanation:
1. P(head) = \( \frac{n}{2n + 1} \cdot 1 + \frac{n + 1}{2n + 1} \cdot \frac{1}{2} = \frac{3n + 1}{2(2n + 1)} \)
2. Set \( \frac{3n + 1}{2(2n + 1)} = \frac{31}{42} \): 42(3n + 1) = 62(2n + 1)
3. 126n + 42 = 124n + 62, so 2n = 20 and n = 10.

Teacher's Note:
1. A two-headed coin shows head with probability 1, a fair coin with probability 1/2.
2. Check: with n = 10, P(head) = 31/42 exactly.

Answer: (a) \( \frac{28}{69} \)

Explanation:
1. Defective branches: A: 0.25 × 0.05 = 0.0125; B: 0.35 × 0.04 = 0.0140; C: 0.40 × 0.02 = 0.0080
2. P(defective) = 0.0345
3. P(B | defective) = \( \frac{0.0140}{0.0345} = \frac{140}{345} = \frac{28}{69} \)

Teacher's Note:
1. Work with percentages as numbers (25 × 5 = 125, 35 × 4 = 140, 40 × 2 = 80) to avoid decimals.
2. The three posteriors should add to 1: 25/69 + 28/69 + 16/69 = 1.

Answer: (a) \( \frac{1}{52} \)

Explanation:
1. Expected accidents: scooter 2000 × 0.01 = 20, car 4000 × 0.03 = 120, truck 6000 × 0.15 = 900.
2. Total = 1040
3. P(scooter | accident) = \( \frac{20}{1040} = \frac{1}{52} \)

Teacher's Note:
1. Using counts instead of priors (2000/12000 etc.) gives the same Bayes ratio with less work.
2. Trucks dominate because both their number and accident rate are highest.

Answer: (d) \( \frac{1}{7} \)

Explanation:
1. k white among 4 transferred: \( P(k) = \frac{^3C_k\,^5C_{4-k}}{^8C_4} \) gives counts 5, 30, 30, 5 (out of 70) for k = 0, 1, 2, 3.
2. P(white | k) = \( \frac{k}{4} \).
3. P(k = 3 | white) = \( \frac{5 \cdot 3}{30 \cdot 1 + 30 \cdot 2 + 5 \cdot 3} = \frac{15}{105} = \frac{1}{7} \)

Teacher's Note:
1. The k = 0 case drops out because it cannot produce a white ball.
2. Common factors like ⁸C₄ and 1/4 cancel, so compare count × k directly.

Answer: (b) \( \frac{24}{29} \)

Explanation:
1. P(guess) = \( \frac{1}{3} \), P(copy) = \( \frac{1}{6} \), P(know) = \( \frac{1}{2} \); P(correct) given each: \( \frac{1}{4}, \frac{1}{8}, 1 \).
2. P(correct) = \( \frac{1}{12} + \frac{1}{48} + \frac{1}{2} = \frac{4 + 1 + 24}{48} = \frac{29}{48} \)
3. P(know | correct) = \( \frac{\frac{24}{48}}{\frac{29}{48}} = \frac{24}{29} \)

Teacher's Note:
1. Find P(know) = 1 − 1/3 − 1/6 = 1/2 first; the three cases are exhaustive.
2. A guess on a four-choice question is correct with probability 1/4.

Answer: (b) \( \frac{7}{19} \)

Explanation:
1. Not meeting standard: A: 0.8 × 0.15 = 0.12; B: 0.2 × 0.35 = 0.07
2. P(not standard) = 0.19
3. P(B | not standard) = \( \frac{0.07}{0.19} = \frac{7}{19} \)

Teacher's Note:
1. The failure rates are 15% and 35%, found as 1 − 85% and 1 − 65%.
2. Using the success rates by mistake would answer a different question.