Q5. Evaluate P(A ∪ B) if 2P(A) = P(B) = 5/13 and P(A|B) = 2/5 (NCERT Exercise 13.1, Q4, page 413).
P(A) = 5/26 and P(A ∩ B) = P(A|B) P(B) = (2/5)(5/13) = 2/13. So P(A ∪ B) = 5/26 + 10/26 − 4/26 = 11/26.
Q6. Two coins are tossed once. Find P(E|F) for (i) E: exactly one coin shows tail, F: exactly one coin shows head; (ii) E: no tail appears, F: no head appears (NCERT Exercise 13.1, Q7, page 414).
The word “one” in NCERT is interpreted as exactly one here. (i) Both E and F are {HT, TH}, so P(E|F) = 1. (ii) E = {HH} and F = {TT} have nothing in common, so P(E|F) = 0.
Q7. A die is thrown three times. Find P(E|F) for E: 4 on the third toss, F: 6 and 5 on the first two tosses (NCERT Exercise 13.1, Q8, page 414).
F has 6 outcomes (6, 5, k), and E ∩ F is only (6, 5, 4): 1/6. The third toss does not depend on the first two.
Q8. Mother, father and son line up at random for a photograph. Find P(E|F) for E: son on one end, F: father in the middle (NCERT Exercise 13.1, Q9, page 414).
F leaves two arrangements, mother–father–son and son–father–mother, and in both the son is at an end: P(E|F) = 1.
Q9. A fair die is rolled, with E = {1, 3, 5}, F = {2, 3} and G = {2, 3, 4, 5}. Find P(E|F), P(F|E), P(E|G), P(G|E), P((E ∪ F)|G) and P((E ∩ F)|G) (NCERT Exercise 13.1, Q11, page 414).
1/2 and 1/3; 1/2 and 2/3; and, since (E ∪ F) ∩ G = {2, 3, 5} and (E ∩ F) ∩ G = {3}, 3/4 and 1/4.
Q10. A die is thrown; if a multiple of 3 comes up it is thrown again, otherwise a coin is tossed. Find the probability that the coin shows a tail given that at least one die shows a 3 (NCERT Exercise 13.1, Q15, page 414).
“At least one die shows a 3” happens only when the die is thrown twice, and then no coin is tossed. So the two events cannot happen together, and the probability is 0.
Q11. If A and B are events with P(A|B) = P(B|A), then (A) A ⊂ B but A ≠ B; (B) A = B; (C) A ∩ B = φ; (D) P(A) = P(B) (NCERT Exercise 13.1, Q17, page 415).
The condition gives P(A ∩ B)/P(B) = P(A ∩ B)/P(A). If P(A ∩ B) > 0, cancellation gives (D), P(A) = P(B), without requiring A = B. But this extra hypothesis is missing from the question. For a fair die, A = {1} and B = {2, 3} have both conditional probabilities 0 while P(A) = 1/6 and P(B) = 1/3. Thus no listed option is universally forced: (D) is the intended answer only with the additional nonzero-intersection assumption.
Q12. A and B are events with P(A) ≠ 0. Find P(B|A) if (i) A is a subset of B; (ii) A ∩ B = φ (NCERT Miscellaneous Exercise on Chapter 13, Q1, page 435).
(i) A ∩ B = A, so P(B|A) = P(A)/P(A) = 1. (ii) P(A ∩ B) = 0, so P(B|A) = 0.
Q13. An electronic assembly has two subsystems A and B, with P(A fails) = 0.2, P(B fails alone) = 0.15 and P(A and B fail) = 0.15. Find (i) P(A fails | B has failed); (ii) P(A fails alone) (NCERT Miscellaneous Exercise on Chapter 13, Q9, page 436).
P(B fails) = 0.15 + 0.15 = 0.3, so (i) 0.15/0.3 = 0.5; (ii) 0.2 − 0.15 = 0.05.
Q14. If P(A|B) > P(A), which is correct? (A) P(B|A) < P(B); (B) P(A ∩ B) < P(A) P(B); (C) P(B|A) > P(B); (D) P(B|A) = P(B) (NCERT Miscellaneous Exercise on Chapter 13, Q12, page 437).
The strict inequality forces P(A) > 0: otherwise P(A ∩ B) and P(A|B) would both be 0. Now P(A|B) > P(A) means P(A ∩ B) > P(A) P(B), so P(B|A) = P(A ∩ B)/P(A) > P(B): (C).