Match List-I with List-I
Let A and B are two events such that $P(A) = 0.8, P(B) = 0.5, P(B|A)= 0.4$
|
List-I |
List-II |
|
(A) $P (A∩ B)$ |
(I) 0.2 |
|
(B) $P(A|B)$ |
(II) 0.32 |
|
(C) $P (A∪B)$ |
(III) 0.64 |
|
(D) $P (A')$ |
(IV) 0.98 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
|
List-I |
List-II |
|
(A) $P (A∩ B)$ |
(II) 0.32 |
|
(B) $P(A|B)$ |
(III) 0.64 |
|
(C) $P (A∪B)$ |
(IV) 0.98 |
|
(D) $P (A')$ |
(I) 0.2 |
$P(A)=0.8,\; P(B)=0.5,\; P(B|A)=0.4$
$P(A\cap B)=P(B|A)\,P(A)$
$=0.4\times0.8$
$=0.32$
$(A)\rightarrow(II)$
$P(A|B)=\frac{P(A\cap B)}{P(B)}$
$=\frac{0.32}{0.5}$
$=0.64$
$(B)\rightarrow(III)$
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
$=0.8+0.5-0.32$
$=0.98$
$(C)\rightarrow(IV)$
$P(A')=1-P(A)$
$=1-0.8$
$=0.2$
$(D)\rightarrow(I)$
Final Matching: (A)-(II), (B)-(III), (C)-(IV), (D)-(I).