Question

If A, B and C are three events such that \[P\left( A \right){\text{ }} = {\text{ }}P\left( B \right){\text{ }} = {\text{ }}P\left( C \right){\text{ }} = {\text{ }}\dfrac{1}{4}{\text{ }},P\left( {AB} \right){\text{ }} = {\text{ }}0\], \[P\left( {AC} \right){\text{ }} = {\text{ }}\dfrac{1}{8}\], then\[\;P\left( {{\text{ }}A{\text{ }} + {\text{ }}B{\text{ }}} \right){\text{ }} = \]
\[\left( 1 \right)\]\[0.125\]
\[\left( 2 \right)\]\[0.25\]
\[\left( 3 \right)\]\[0.375\]
\[\left( 4 \right)\]\[0.5\]

Answer 1

Hint: We have to find the\[P\left( {{\text{ }}A{\text{ }} + {\text{ }}B{\text{ }}} \right)\]. We solve the question using condition probability and using the concept of meaning of some formulas . We also use the knowledge of the identity of probability of union of two events . From the given data we can compute the values of the probability of \[P{\text{ }}\left( {{\text{ }}A{\text{ }} + {\text{ }}B{\text{ }}} \right){\text{ }}.\]

Complete step-by-step solution:
The intersection of terms are said to the common elements shared by the two events . The intersection is also stated as “ and “ . whereas the union of the terms is said to be the sum total of the two events and subtracting the common portion or the intersection of the two events . The union is also stated as “ or “ .
Given :
\[P\left( A \right){\text{ }} = {\text{ }}P\left( B \right){\text{ }} = {\text{ }}P\left( C \right){\text{ }} = {\text{ }}\dfrac{1}{4}{\text{ }},{\text{ }}P\left( {AB} \right){\text{ }} = {\text{ }}0{\text{ }},{\text{ }}P\left( {AC} \right){\text{ }} = {\text{ }}\dfrac{1}{8}\]
Recognising,
\[P\left( {AB} \right){\text{ }} = {\text{ }}P\left( {{\text{ }}A{\text{ }}|{\text{ }}B{\text{ }}} \right)\]
And we know that
\[P\left( {{\text{ }}A{\text{ }}|{\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}\dfrac{{P{\text{ }}\left( {{\text{ }}A{\text{ }} \cap B{\text{ }}} \right){\text{ }}}}{{{\text{ }}P{\text{ }}\left( {{\text{ }}B{\text{ }}} \right)}}\]
Now ,
\[P\left( {AB} \right){\text{ }} = \dfrac{{P\left( {{\text{ }}A{\text{ }} \cap {\text{ }}B{\text{ }}} \right)}}{{{\text{ }}P\left( {{\text{ }}B{\text{ }}} \right)}}\]
As given , \[P\left( {AB} \right){\text{ }} = {\text{ }}0\]
So ,
\[\dfrac{{P\left( {{\text{ }}A{\text{ }} \cap {\text{ }}B{\text{ }}} \right)}}{{P\left( {{\text{ }}B{\text{ }}} \right)}}{\text{ }} = {\text{ }}0\;\]
From here we can say that
\[P\left( {{\text{ }}A{\text{ }} \cap {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}0\]
We know that the probability of sum of two events is given by the formula :
\[P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}P\left( A \right){\text{ }} + {\text{ }}P\left( B \right){\text{ }} - {\text{ }}P\left( {{\text{ }}A{\text{ }} \cap {\text{ }}B{\text{ }}} \right)\]
Putting , value of \[P\left( {{\text{ }}A{\text{ }} \cap {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}0\]
\[P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}P\left( A \right){\text{ }} + {\text{ }}P\left( B \right)\]
\[\Rightarrow P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}1/4{\text{ }} + {\text{ }}1/4\]
\[\Rightarrow P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}1/2\]
\[\Rightarrow P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}0.5\]
\[\Rightarrow P\left( {{\text{ }}A{\text{ }} \cup {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}P\left( {{\text{ }}A{\text{ }} + {\text{ }}B{\text{ }}} \right)\]
As , both have the same meaning
So ,
\[P\left( {{\text{ }}A{\text{ }} + {\text{ }}B{\text{ }}} \right){\text{ }} = {\text{ }}0.5\]
Thus , the correct option is \[\left( 4 \right)\]

Note: We don’t need the values of \[P\left( C \right)\]and \[P\left( {AC} \right)\]for solving the required problem . It was just to create a sense of confusion . Using the data we can calculate the value \[of{\text{ }}P\left( {{\text{ }}A{\text{ }} + {\text{ }}C{\text{ }}} \right)\]as we can evaluate the value of \[P\left( {A{\text{ }} \cap {\text{ }}C} \right)\]from the formula stated above .
We find the relation by using the conditional probability formula . If $A$ and $B$are mutually exclusive events , then\[P\left( {A{\text{ }} \cup {\text{ }}B} \right){\text{ }} = {\text{ }}P\left( A \right){\text{ }} + {\text{ }}P\left( B \right)\]. The basic property of probability is that the probability of an event can never be greater than$1$.
If two events $A$ and $B$ are independent , then
\[P\left( {E{\text{ }} \cap {\text{ }}F} \right){\text{ }} = P\left( E \right){\text{ }} \times {\text{ }}P\left( F \right)\]
\[P\left( {E{\text{ }}|{\text{ }}F} \right){\text{ }} = {\text{ }}P\left( E \right){\text{ }},{\text{ }}P\left( F \right){\text{ }} \ne 0\]
\[P\left( {F{\text{ }}|{\text{ }}E} \right){\text{ }} = {\text{ }}P\left( F \right){\text{ }},{\text{ }}P\left( E \right){\text{ }} \ne {\text{ }}0\]