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 solve the problem of matching List-I (probability events related to coin tosses) with List-II (values of probabilities), let's calculate each probability:

• (A) P(exactly 2 heads):
  The sample space when a coin is tossed twice is: {HH, HT, TH, TT}. The event "exactly 2 heads" is {HH}. Thus, P(A) = \(\frac{1}{4}\), which matches with (I).
• (B) P(at least 1 head):
  The favorable outcomes are {HH, HT, TH}. Thus, P(B) = \(\frac{3}{4}\), which matches with (III).
• (C) P(at most 2 heads):
  This includes all outcomes in the sample space {HH, HT, TH, TT}, thus P(C) = \(1\), which matches with (II).
• (D) P(exactly 1 head):
  The favorable outcomes are {HT, TH}. Thus, P(D) = \(\frac{1}{2}\), which matches with (IV).

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

Answer 2

To solve the problem, we need to calculate the probabilities for each event listed in List-I when a coin is tossed twice. The possible outcomes of tossing a coin twice are: HH, HT, TH, TT. These represent two heads, a head followed by a tail, a tail followed by a head, and two tails. 

1. P(exactly 2 heads): From the outcomes (HH, HT, TH, TT), only HH has exactly 2 heads. Therefore, P(exactly 2 heads) = 1/4.
2. P(at least 1 head): At least 1 head occurs in the outcomes HH, HT, and TH. Therefore, P(at least 1 head) = 3/4.
3. P(at most 2 heads): Since 2 heads is the maximum possible in two coin tosses, all outcomes (HH, HT, TH, TT) are considered, giving P(at most 2 heads) = 1.
4. P(exactly 1 head): The outcomes HT and TH have exactly 1 head each. Therefore, P(exactly 1 head) = 2/4 = 1/2.

Now, match these probabilities with List-II:

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

Therefore, the correct matching is: (A)-(I), (B)-(III), (C)-(II), (D)-(IV)