Question

For any two events $A$ and $B$ in a sample space
A. ${\rm P}\left( {\dfrac{{\rm A}}{{\rm B}}} \right) \geqslant \dfrac{{{\rm P}({\rm A}) + {\rm P}({\rm B}) - 1}}{{{\rm P}({\rm B})}},{\rm P}({\rm B}) \ne 0$ , is always true.
B. ${\rm P}({\rm A} \cap {\rm B}) = {\rm P}({\rm A}) - {\rm P}(\overline {\rm A} \cap \overline {{\rm B})}$ does not hold
C. ${\rm P}({\rm A} \cup {\rm B}) = 1 - {\rm P}(\overline {\rm A} ){\rm P}(\overline {\rm B} )$ , if ${\rm A}$ and ${\rm B}$ are independent
D. ${\rm P}({\rm A} \cup {\rm B}) = 1 - {\rm P}({\rm A}){\rm P}({\rm B})$ , if ${\rm A}$ and ${\rm B}$ are disjoint.

Answer 1

Hint : Here, any two events $A$ and $B$ in sample space. The sample space of a random experiment is the collection of all possible outcomes. A probability model consists of the sample space and the way to assign probabilities. An event associated with a random experiment is a subset of the sample space. The probability of any outcome is a number between 0 and 1. An event is a set of outcomes of an experiment to which a probability is assigned.

Complete step by step solution:
${\rm P}\left( {\dfrac{{\rm A}}{{\rm B}}} \right) \geqslant \dfrac{{{\rm P}({\rm A}) + {\rm P}({\rm B}) - 1}}{{{\rm P}({\rm B})}},{\rm P}({\rm B}) \ne 0$ , is always true.
We know that, ${\rm P}\left( {\dfrac{{\rm A}}{{\rm B}}} \right) = \dfrac{{{\rm P}({\rm A} \cap {\rm B})}}{{{\rm P}(B)}}$ where ${\rm P}({\rm B}) \ne 0$
Now,
${\rm P}({\rm A} \cap {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A} \cup {\rm B})$
Rearranging it, we get
${\rm P}({\rm A} \cup {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A} \cap {\rm B})$
By using above values, we have

$${\rm P}\left( {\dfrac{{\rm A}}{{\rm B}}} \right) = \dfrac{{{\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A} \cup {\rm B})}}{{{\rm P}(B)}} \\\ {\rm P}\left( {\dfrac{{\rm A}}{{\rm B}}} \right) \geqslant \dfrac{{{\rm P}({\rm A}) + {\rm P}({\rm B}) - 1}}{{{\rm P}(B)}} \;$$

${\rm P}({\rm A} \cup {\rm B}) \leqslant 1$
Hence, Option(A) is true

Therefore, ${\rm P}(\overline {\rm A} \cap \overline {\rm B} ) = 1 - {\rm P}({\rm A} \cup B)$ , By apply demorgan’s law into the above equation, we get,
Here, we have
${\rm P}({\rm A} \cap {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A} \cup {\rm B})$
By this demorgan’s law, we get
${\rm P}({\rm A} \cap {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - 1 + {\rm P}(\overline {\rm A} \cap \overline {\rm B} )$
Hence, Option(B) is also true.

So we use ${\rm P}({\rm A} \cup {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A} \cap {\rm B})$
if ${\rm A}$ and ${\rm B}$ are independent, then ${\rm P}({\rm A} \cap {\rm B}) = {\rm P}({\rm A}){\rm P}({\rm B})$
Thus, ${\rm P}({\rm A} \cup {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) - {\rm P}({\rm A}){\rm P}({\rm B})$
Take ${\rm P}({\rm A})$ out from bracket, we get

$${\rm P}({\rm A} \cup {\rm B}) = {\rm P}({\rm A})(1 - {\rm P}({\rm B})) + {\rm P}({\rm B}) \\\ = (1 - {\rm P}(\overline {\rm A} )){\rm P}(\overline {\rm B} ) + 1 - {\rm P}(\overline {\rm B} ) \;$$

Therefore, ${\rm P}({\rm A} \cup {\rm B}) = 1 - {\rm P}(\overline {\rm A} ){\rm P}(\overline {\rm B} )$ , which is option(C)

if ${\rm A}$ and ${\rm B}$ are disjoint, then ${\rm P}({\rm A} \cap {\rm B}) = 0$
We know, ${\rm P}({\rm A} \cup {\rm B}) = {\rm P}({\rm A}) + {\rm P}({\rm B}) = 2 - {\rm P}(\overline {\rm A} ) - {\rm P}(\overline {\rm B} )$ ,
Finally, ${\rm A}$ , ${\rm B}$ and $C$ is the final answer.
So, the correct answer is “OptionA,B and C”.

Note : In probability theory, an event is a set of outcomes of an experiment to which a probability is assigned. A single outcome may be an element of many different events, and different events in an experiment are usually not equally likely, since they may include very different groups of outcomes.