Question

If $A$ and $B$ are two independent events such that $P\left( {A \cup B'} \right) = 0.8$ and $P\left( A \right) = 0.3$. Then $P\left( B \right)$ is equal to
A. $\dfrac{2}{7}$
B. $\dfrac{2}{3}$
C. $\dfrac{3}{8}$
D. $\dfrac{1}{4}$

Answer 1

Hint: Independent events are those events whose occurrence is not dependent on any other event. We know that for any two events $A$ and $B$, $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$, we have given that $A$ and $B$ are two independent events i.e., $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$.
At first, we will apply $P\left( {A \cup B'} \right) = P\left( A \right) + P\left( {B'} \right) - P\left( A \right)P\left( {B'} \right)$ formula and calculate the value of $P\left( {A \cup B'} \right) = P\left( A \right) + P\left( {B'} \right) - P\left( A \right)P\left( {B'} \right)$. After this, we will apply the $P\left( X \right) = 1 - P\left( {X'} \right)$ formula to get the value of $P\left( B \right)$.

Formula Used:
$P\left( {A \cup B'} \right) = P\left( A \right) + P\left( {B'} \right) - P\left( A \right)P\left( {B'} \right)$
If $X$ is any event, then $P\left( X \right) = 1 - P\left( {X'} \right)$.

Complete step by step solution:
Given that, $P\left( {A \cup B'} \right) = 0.8$ and $P\left( A \right) = 0.3$
As we know that $P\left( {A \cup B'} \right) = P\left( A \right) + P\left( {B'} \right) - P\left( A \right)P\left( {B'} \right)$. Therefore, on substituting the given values in this formula, we get
\[ \Rightarrow 0.8 = 0.3 + P\left( {B'} \right) - 0.3 \times P\left( {B'} \right)\]
\[ \Rightarrow 0.8 - 0.3 = P\left( {B'} \right) - 0.3P\left( {B'} \right)\]
Now, take $P\left( {B'} \right)$ as a common term
\[ \Rightarrow 0.8 - 0.3 = P\left( {B'} \right)\left( {1 - 0.3} \right)\]
On subtraction, we get
\[ \Rightarrow 0.5 = 0.7P\left( {B'} \right)\]
Now, divide the above written equation by $0.7$.
\[ \Rightarrow P\left( {B'} \right) = \dfrac{{0.5}}{{0.7}}\]
\[ \Rightarrow P\left( {B'} \right) = \dfrac{5}{7}\]
We know that $P\left( B \right) = 1 - P\left( {B'} \right)$. On substituting the value of $P\left( {B'} \right)$ in $P\left( B \right) = 1 - P\left( {B'} \right)$ formula, we get
$ \Rightarrow P\left( B \right) = 1 - \dfrac{5}{7}$
$ \Rightarrow P\left( B \right) = \dfrac{{7 - 5}}{7}$
On subtraction, we get
$\therefore P\left( B \right) = \dfrac{2}{7}$
Hence, the value of $P\left( B \right)$ is $\dfrac{2}{7}$.

Option ‘A’ is correct

Note: Always remember that for any event $X$, $P\left( X \right) = 1 - P\left( {X'} \right)$ where $P\left( {X'} \right)$ is $P\left( {not\,X} \right)$ and for any two independent events $A$ and $B$, $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$. Also, remember that if $A$ and $B$ are mutually exclusive events, then $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$. Remember that independent events can have common outcomes but mutually exclusive events can never have common outcomes. Students should take care of the calculations so as to be sure of our final answer.