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

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
$P\left( {E{\text{ }} \cap {\text{ }}F} \right){\text{ }} = P\left( E \right){\text{ }} \times {\text{ }}P\left( F \right)$
$P\left( {E{\text{ }}|{\text{ }}F} \right){\text{ }} = {\text{ }}P\left( E \right){\text{ }},{\text{ }}P\left( F \right){\text{ }} \ne 0$
$P\left( {F{\text{ }}|{\text{ }}E} \right){\text{ }} = {\text{ }}P\left( F \right){\text{ }},{\text{ }}P\left( E \right){\text{ }} \ne {\text{ }}0$