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.