A coin is tossed twice. Then
Match List-I with List-II:
|
List-I |
List-II |
|
(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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
|
List-I |
List-II |
|
(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) $\frac{1}{2}$ |
$\text{Sample space }=\{HH,HT,TH,TT\}.$
$(A)\;P(\text{exactly 2 heads})=\frac{1}{4}\Rightarrow(I).$
$(B)\;P(\text{at least 1 head})=\frac{3}{4}\Rightarrow(III).$
$(C)\;P(\text{at most 2 heads})=1\Rightarrow(II).$
$(D)\;P(\text{exactly 1 head})=\frac{2}{4}=\frac{1}{2}\Rightarrow(IV).$
$(A)\rightarrow(I),\;(B)\rightarrow(III),\;(C)\rightarrow(II),\;(D)\rightarrow(IV).$