Question

Two events A and B will be independent if
A. P(A’$ \cap $B’) = (1- P(A)) (1-P(B)).
B. P(A) + P(B) = 1
C. P(A) = P(B)
D. A and B are mutually exclusive.

Answer 1

Hint: The strategy for solving this question is that first of all use the definition of independent events to write the condition for two events to be independent and then use the relation between the number and its complement to get the answer.

Complete step-by-step answer:
First of all we will see the definition of independent events.
Independent events are those events which are not affected by previous events.
For example: A coin does not know it came up heads before. So coming of head or tail next time does not depend upon the previous events.
The formula for finding conditional probability is given as:
P (A/B) = $\dfrac{{{\text{P(A}} \cap B)}}{{{\text{P(A)}}}}$ .
If the two events A and B are independent then P (A/B) = P(A).
Therefore the condition for two events A and B to be independent is given as:
P(A) P(B) = P(A$ \cap $B).
Now, the option A can be written as:
P(A’$ \cap $B’)=P(A’) P(B’). (1)
We know the relation between the event A and its complement is given as:
P(A) + P(A’) = 1
P (A’) = 1- P(A).
Similarly, we can write:
P (B’) = 1- P (B).
Putting the above values in equation 1, we get:
P (A’$ \cap $B’) = (1-P(A)) (1- P(B)).
So option A is correct.

Note: For solving this type of question, we should remember the basic concept of independent events i.e. its definition and the condition for the two events to be independent. We should note that two events A and B are said to be mutually exclusive if it is not possible that both of them occur at same time. So it is not the same as independent events.