If A and B are two independent events then,
Match List I with List II
| LIST I | LIST II | ||
| A. | \(P(A\cup B)\) | I. | \(P(A')P(B')\) |
| B. | P(not A or not B) | II. | \(P(A)\); \(P(B)\ne 0\) |
| C. | P(A/B) | III. | \(1-P(A')P(B')\) |
| D. | P(neither A nor B) | IV. | \(1-P(A)P(B)\) |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-II, D-I
For independent events, \(P(A\cap B)=P(A)P(B)\), and A′ and B′ are also independent, so \(P(A'\cap B')=P(A')P(B')\).
(A) \(P(A\cup B)\)
This matches (III) \(1-P(A')P(B')\). The complement of “A or B” is “neither A nor B”:
\(P(A\cup B)=1-P(A'\cap B')=1-P(A')P(B')\)
(B) P(not A or not B)
This matches (IV) \(1-P(A)P(B)\). The complement of “not A or not B” is “both A and B”:
\(P(A'\cup B')=1-P(A\cap B)=1-P(A)P(B)\)
(C) P(A/B)
This matches (II) P(A), with \(P(B)\ne 0\):
\( \displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)P(B)}{P(B)}=P(A) \)
(D) P(neither A nor B)
This matches (I) \(P(A')P(B')\), since \(P(A'\cap B')=P(A')P(B')\).