Question

: If A and B are mutually exclusive events with \[P(B) \ne 1\], then \[P(\dfrac{A}{{{B'}}}) = \]
1) \[\dfrac{1}{{P(B)}}\]
2) \[\dfrac{1}{{1 - P(B)}}\]
3) \[\dfrac{{P(A)}}{{P(B)}}\]
4) \[\dfrac{{P(A)}}{{1 - P(B)}}\]

Answer 1

Hint: We are given two events which are mutually exclusive and also given that, \[P(B) \ne 1\]. We need to find the value of \[P(\dfrac{A}{{{B'}}})\] . We know that, for mutually exclusive events A and B, \[P(A \cap B) = 0\] . Also, we know that, \[P(B) + P({B'}) = 1\] and \[P(A \cap {B'}) = P(A)\]. Thus, we will use this and get the final output.

Complete step-by-step answer:
We know that,
In probability theory, two events are said to be mutually exclusive if they cannot occur at the same time or simultaneously. This means that, if two events are considered mutually exclusive events, then the probability of both events occurring at the same time will be zero.
Given that, A and B are mutually exclusive events.
\[\therefore P(A \cap B) = 0\] ---- (1)
We know that,
For event B, the complement of this event is B’
\[P(B) + P({B'}) = 1\]
\[ \Rightarrow P({B'}) = 1 - P(B)\] ---- (2)
Now we will find,
\[P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A \cap {B'})}}{{P({B'})}}\]
We know that, \[P(A \cap {B'}) = P(A)\]
Substitute this, we will get,
\[ \Rightarrow P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A) - P(A \cap B)}}{{P({B'})}}\]
Substitute the value of equation (1), we will get,
\[ \Rightarrow P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A) - 0}}{{P({B'})}}\]
Substitute the value of equation (2), we will get,
\[ \Rightarrow P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A)}}{{1 - P(B)}}\]
So, the correct answer is “Option 4”.

Note: Another Method:
\[P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A \cap {B'})}}{{P({B'})}}\]
We know that, \[P(A \cap {B'}) = P(A)\]
Substitute this value, we will get,
\[ \Rightarrow P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A)}}{{P({B'})}}\]
Substitute the value of equation (2), we will get,
\[ \Rightarrow P(\dfrac{A}{{{B'}}}) = \dfrac{{P(A)}}{{1 - P(B)}}\]