A coin is tossed twice. Then
Match List-I with List-II:
| List-I | List-II | ||
| (A) | P(exactly 2 heads) | (I) | \(\displaystyle\frac{1}{4}\) |
| (B) | P(at least 1 head) | (II) | 1 |
| (C) | P(at most 2 heads) | (III) | \(\displaystyle\frac{3}{4}\) |
| (D) | P(exactly 1 head) | (IV) | \(\displaystyle\frac{1}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
When a coin is tossed twice, the sample space has 4 equally likely outcomes:
\(S = \{HH, HT, TH, TT\}\)
(A) P(exactly 2 heads)
This matches (I) 1/4. Only HH has two heads:
\( \displaystyle P = \frac{1}{4} \)
(B) P(at least 1 head)
This matches (III) 3/4. Every outcome except TT has a head:
\( \displaystyle P = 1 - \frac{1}{4} = \frac{3}{4} \)
(C) P(at most 2 heads)
This matches (II) 1. Two tosses can never give more than 2 heads, so this is a sure event:
\(P = 1\)
(D) P(exactly 1 head)
This matches (IV) 1/2. The outcomes are HT and TH:
\( \displaystyle P = \frac{2}{4} = \frac{1}{2} \)
So the matching is (A)-(I), (B)-(III), (C)-(II), (D)-(IV).