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

Answer: (A)

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

Answer 2

(A): $P(exactly 2 heads)$ occurs when both tosses are heads (HH). The probability is:
$P(exactly 2 heads) = \frac{1}{4}$.
Thus, (A)-(I).
(B): $P(at least 1 head)$ occurs in the cases HT, TH, and HH. The probability is:
$P(at least 1 head) = 1 - P(no heads) = 1 - \frac{1}{4} = \frac{3}{4}$.
Thus, (B)-(III).
(C): $P(at most 2 heads)$ includes all possible outcomes (HH, HT, TH, TT). The probability is:
$P(at most 2 heads) = 1$.
Thus, (C)-(II).
(D): $P(exactly 1 head)$ occurs in the cases HT and TH. The probability is:
$P(exactly 1 head) = \frac{2}{4} = \frac{1}{2}$.
Thus, (D)-(IV).