If A and B are independent events and $P(A) =\frac{1}{2}, P(B) =\frac{1}{3}$ then
Match List-I with List-II
|
List-I |
List-II |
|
(A) $P(A∩B)$ |
(I) $\frac{1}{2}$ |
|
(B) $P(\bar A)P(B) + P(A)P(\bar B)$ |
(II) $\frac{1}{3}$ |
|
(C) $P(A|B) + P(B | A)$ |
(III) $\frac{1}{6}$ |
|
(D) $P(A∩\bar B)$ |
(IV) $\frac{5}{6}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
|
List-I |
List-II |
|
(A) $P(A∩B)$ |
(III) $\frac{1}{6}$ |
|
(B) $P(\bar A)P(B) + P(A)P(\bar B)$ |
(I) $\frac{1}{2}$ |
|
(C) $P(A|B) + P(B | A)$ |
(IV) $\frac{5}{6}$ |
|
(D) $P(A∩\bar B)$ |
(II) $\frac{1}{3}$ |
Given: $P(A)=\frac{1}{2},\ P(B)=\frac{1}{3}$ and A, B are independent.
Then $P(A\cap B)=P(A)P(B)=\frac{1}{6}$.
(A) $P(A\cap B)=\frac{1}{6}$ → (III)
(B) $P(\bar{A})P(B)+P(A)P(\bar{B}) = (1-\frac{1}{2})(\frac{1}{3}) + (\frac{1}{2})(1-\frac{1}{3}) = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ → (I)
(C) $P(A|B)+P(B|A) = \frac{P(A\cap B)}{P(B)} + \frac{P(A\cap B)}{P(A)} = \frac{\frac{1}{6}}{\frac{1}{3}} + \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ → (IV)
(D) $P(A\cap\bar{B}) = P(A) - P(A\cap B) = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ → (II)