Single correct option
Match the experiments in List-I with their probabilities in List-II.
List-I
(P) A fair die is thrown three times. The probability that the three outcomes form a strictly increasing sequence.
(Q) A bag has 3 red and 2 blue balls. Two balls are drawn without replacement. The probability that both are red, given that at least one is red.
(R) A fair coin is tossed repeatedly until two consecutive heads appear for the first time. The probability that this happens exactly at the fourth toss.
(S) The probability that a randomly chosen permutation of has no element in its natural position.
List-II
(1)
(2)
(3)
(4)
(5)
List-I
(P) A fair die is thrown three times. The probability that the three outcomes form a strictly increasing sequence.
(Q) A bag has 3 red and 2 blue balls. Two balls are drawn without replacement. The probability that both are red, given that at least one is red.
(R) A fair coin is tossed repeatedly until two consecutive heads appear for the first time. The probability that this happens exactly at the fourth toss.
(S) The probability that a randomly chosen permutation of has no element in its natural position.
List-II
(1)
(2)
(3)
(4)
(5)
- A. P→4, Q→3, R→1, S→2
- B. P→4, Q→5, R→2, S→1
- C. P→4, Q→5, R→1, S→2
- D. P→1, Q→5, R→4, S→2
Show answer and solution
Answer: (C) P→4, Q→5, R→1, S→2
(P) A strictly increasing triple needs three distinct values chosen from six, so there are favourable outcomes out of , giving , which is (4). (Q) and , so the conditional probability is , which is (5). (R) The game ends at toss 4 only for sequences whose last two tosses are HH and which have no earlier HH; the second toss must therefore be T (otherwise tosses 2 and 3 would already be HH), leaving HTHH and TTHH, so the probability is , which is (1). (S) The number of derangements of five objects is , so the probability is , which is (2). The value (3) is a distractor.
Saved on this device. Sign in or create a free account to keep it on every device.