Question

Let $A$ and $B$ are two events and $P({{A}^{'}})=0.3$, $P(B)=0.4$, $P(A\cap {{B}^{'}})=0.5$, then $P(A\cup {{B}^{'}})$ is

Answer 1

Hint: In this question, we are to find the probability of an event. For this question, the addition theorem on probability is used. All the given values are substituted in the addition theorem of probability to find the required probability.

Formula used: A probability is the ratio of favorable outcomes of an event to the total number of outcomes. So, the probability lies in between \[0\] and \[1\].
The probability is calculated by,
$P(E)=\dfrac{n(E)}{n(S)}$
$n(E)$- favourable outcomes and $n(S)$ - sample.
If there are two events in a sample space, then the addition theorem on probability is given by
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$
When two events happen independently, the occurrence of one is not impacted by the occurrence of the other.
For the events $A$ and $B$, $P(A\cap B)=P(A)\cdot P(B)$ if they are independent and $P(A\cap B)=\Phi $ if they are mutually exclusive.
The probability of a complementary event:
$P({{A}^{'}})=1-P(A)$

Complete step by step solution: Given that events $A$ and $B$ are two,
$P({{A}^{'}})=0.3$
$P(B)=0.4$
$P(A\cap {{B}^{'}})=0.5$
Then, by addition theorem on probability,
$P(A\cup {{B}^{'}})=P(A)+P({{B}^{'}})-P(A\cap {{B}^{'}})$
Here,
$\begin{align}
  & P(A)=1-P({{A}^{'}}) \\
 & \text{ }=1-0.3 \\
 & \text{ }=0.7 \\
\end{align}$
$\begin{align}
  & P({{B}^{'}})=1-P(B) \\
 & \text{ }=1-0.4 \\
 & \text{ }=0.6 \\
\end{align}$
On substituting,
$\begin{align}
  & P(A\cup {{B}^{'}})=P(A)+P({{B}^{'}})-P(A\cap {{B}^{'}}) \\
 & \text{ }=0.7+0.6-0.5 \\
 & \text{ }=0.13-0.5 \\
 & \text{ }=0.8 \\
\end{align}$

Thus, Option (B) is correct.

Note: In this question, the addition theorem on probability is applied for finding the required probability. By substituting the appropriate values, the required probability is calculated. Here we may go wrong with the complimented probability. The actual probability must be calculated.