Probability: JEE Advanced PYQ, 2007–2026

Answer: (D)

There are C(6, 3) = 20 sequences with exactly three heads, each of probability (1/2)⁶ = 1/64. The probability is 20/64 = 5/16.

Answer: (A)

Treat the hand as an unordered set of 4 cards, so there are C(52,4) = 270725 equally likely hands. A hand with all four suits has one card from each suit, which can be chosen in 13 × 13 × 13 × 13 = 28561 ways. So P = 28561/270725 = 2197/20825, which is about 0.105.

Answer: (A)

By the addition theorem, P(A ∩ B) = P(A) + P(B) − P(A ∪ B). Since P(A ∪ B) ≤ 1, P(A ∩ B) ≥ 3/4 + 5/8 − 1 = 3/8. Also A ∩ B ⊂ B, so P(A ∩ B) ≤ P(B) = 5/8, and this is smaller than P(A) = 3/4, so it is the binding upper bound. Both extremes can occur: P(A ∩ B) = 5/8 when B ⊂ A, and P(A ∩ B) = 3/8 when A ∪ B is the whole sample space. Then P(A | B) = P(A ∩ B)/(5/8) lies between (3/8)/(5/8) = 3/5 and (5/8)/(5/8) = 1.

Answer: (B)

366 = 52 × 7 + 2. The two extra days are consecutive, and can be any of 7 equally likely pairs; a Sunday is among them in 2 of the pairs (Saturday–Sunday and Sunday–Monday). The probability is 2/7.

Answer: (A)

Only the positions of the S's matter, and each set of 4 positions out of 11 is equally likely, so there are C(11,4) = 330 outcomes. The S's are consecutive if they start at position 1, 2, ..., 8, which gives 8 favourable outcomes. So P = 8/330 = 4/165. Counting arrangements gives the same result: 8!/(4!·2!) = 840 favourable out of 11!/(4!·4!·2!) = 34650, and 840/34650 = 4/165.

Answer: (A)

Use Bayes' theorem with the two test results combined into one event. Given the disease status, the tests are independent, so P(++|D) = 0.9² = 0.81 and P(++|Dᶜ) = 0.05² = 0.0025. Then P(D|++) = (0.02·0.81)/(0.02·0.81 + 0.98·0.0025) = 0.0162/0.01865 = 324/373. Applying the prior only once and squaring both likelihoods is the key step. The result is not the single-test value 18/67, and it is not found by dropping the prior.

Answer: (A)

From P(A | B) = P(A ∩ B)/P(B), P(B) = (1/12)/(1/2) = 1/6. From P(B | A) = P(A ∩ B)/P(A), P(A) = (1/12)/(1/3) = 1/4. The addition theorem gives P(A ∪ B) = 1/4 + 1/6 − 1/12 = 3/12 + 2/12 − 1/12 = 4/12 = 1/3.

Answer: (A)

P(X = 1) = 2/5. P(X = 2) = (3/5)(2/4) = 3/10. P(X = 3) = (3/5)(2/4)(2/3) = 1/5. P(X = 4) = (3/5)(2/4)(1/3)(2/2) = 1/10. These sum to 1. E(X) = 2/5 + 6/10 + 3/5 + 4/10 = 2. E(X²) = 2/5 + 12/10 + 9/5 + 16/10 = 5. Var(X) = 5 − 2² = 1.

Answer: (A)

The sample space is {HH, HT, TH, TT}. P(A) = P(B) = P(C) = 1/2. The pairwise intersections are A ∩ B = {HH}, A ∩ C = {HH} and B ∩ C = {HH}, each with probability 1/4 = (1/2)(1/2), so the events are pairwise independent. The triple intersection A ∩ B ∩ C = {HH} has probability 1/4, while P(A)P(B)P(C) = 1/8. So the events are not mutually independent.

Answer: (A)

For k white balls, P(both white) = C(k, 2)/C(4, 2) = C(k, 2)/6. This is 0 for k = 0, 1 and it is 1/6, 3/6, 6/6 for k = 2, 3, 4. As the k are equally likely beforehand, Bayes' theorem gives updated probabilities 1/10, 3/10, 6/10. With two whites already drawn, the chance that the third is white is 0, 1/2, 1 for k = 2, 3, 4. So the answer is (3/10)(1/2) + (6/10)(1) = 3/4. As a second way, compare the whole events. P(first two white) = (1/5)(1/6 + 3/6 + 6/6) = 1/3. P(first three white) = (1/5)(1/4 + 1) = 1/4, because with 3 whites the first three draws are all white with probability 1/C(4, 3) = 1/4, and with 4 whites they always are. The ratio is (1/4)/(1/3) = 3/4.