Question

If $P\left( A \right) = 0.40,P\left( B \right) = 0.35$ and $P\left( {A \cup B} \right) = 0.55$, the $P\left( {A/B} \right) = $
A) $\dfrac{1}{5}$
B) $\dfrac{8}{{11}}$
C) $\dfrac{4}{7}$
D) $\dfrac{3}{4}$

Answer 1

Hint: Here, we have to find the conditional probability for A given B. The formula for finding the conditional probability is
$ P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$
But, here we are given $P\left( {A \cup B} \right)$ instead of $P\left( {A \cap B} \right)$. So, we will use the following formula to find $P\left( {A \cap B} \right)$.
$ P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

Complete step by step solution:
In this question, we are given the probability of two events A and B and the probability of $A \cup B$, and we need to find the conditional probability of A given B.
Given data:
$P\left( A \right) = 0.40$
$P\left( B \right) = 0.35$
$P\left( {A \cup B} \right) = 0.55$
Now, first of all let us see what conditional probability is.
If we are given two events A and B, in a sample space S, then the conditional probability of A given B is defined as
$ \Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$, where $P\left( B \right) > 0$
So, to find the conditional probability, we need two parameters: $P\left( {A \cap B} \right)$ and $P\left( B \right)$.
Here, we have $P\left( B \right)$ but we do not have $P\left( {A \cap B} \right)$. Instead we have $P\left( {A \cup B} \right)$. Now, we know the definition that union means a common set including all the elements of set A and set B excluding the common elements. So, therefore, we could write
$
   \Rightarrow P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \\
   \Rightarrow 0.55 = 0.40 + 0.35 - P\left( {A \cap B} \right) \\
   \Rightarrow P\left( {A \cap B} \right) = 0.75 - 0.55 \\
   \Rightarrow P\left( {A \cap B} \right) = 0.20 \\
 $
Hence, now we can find the conditional probability of A given B.
$
   \Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} \\
   \Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{{0.20}}{{0.35}} \\
   \Rightarrow P\left( {\dfrac{A}{B}} \right) = \dfrac{20}{35}=\dfrac{4}{7} \\
 $
Therefore, $P\left( {A/B} \right) = \dfrac{4}{7}$. So, Option (C), $\dfrac{4}{7}$ is the correct answer.

Note:
Axiom 1: For an event A, $P\left( {\dfrac{A}{B}} \right) \geqslant 0$.
Axiom 2: Conditional probability of B given B is always equal to 1. That is
$P\left( {\dfrac{B}{B}} \right) = 1$
Axiom 3: If A, B and C are disjoint events, then
$P\left( {\dfrac{{A \cup B \cup C}}{E}} \right) = P\left( {\dfrac{A}{E}} \right) + P\left( {\dfrac{B}{E}} \right) + P\left( {\dfrac{C}{E}} \right)$