Question

A coin is tossed twice. Then Match List-I with List-II:

List-I	List-II (Adverbs)
(A) P(exactly 2 heads)	(I) \(\frac{1}{4}\)
(B) P(at least 1 head)	(II) \(1\)
(C) P(at most 2 heads)	(III) \(\frac{3}{4}\)
(D) P(exactly 1 head)	(IV) \(\frac{1}{2}\)

Choose the correct answer from the options given below :

• A. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
• B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
• C. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
• D. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Answer 1

The Correct Option is A

To match List-I (probability events from coin tosses) with List-II (probability values), we calculate each probability:

• (A) P(exactly 2 heads):
  For two coin tosses, the sample space is {HH, HT, TH, TT}. The event "exactly 2 heads" is {HH}. Therefore, P(A) = \(\frac{1}{4}\), matching (I).
• (B) P(at least 1 head):
  The favorable outcomes are {HH, HT, TH}. Thus, P(B) = \(\frac{3}{4}\), matching (III).
• (C) P(at most 2 heads):
  This includes all outcomes in the sample space {HH, HT, TH, TT}. Therefore, P(C) = \(1\), matching (II).
• (D) P(exactly 1 head):
  The favorable outcomes are {HT, TH}. Thus, P(D) = \(\frac{1}{2}\), matching (IV).

The correct matches are: (A)-(I), (B)-(III), (C)-(II), (D)-(IV).