If E and F be events in a sample space such that \[P(E \cup F) = 0.8\],\[P\left( {E \cap F} \right) = 0.3\]and\[P\left( E \right) = 0.5\], then \[P\left( F \right)\] is
A) \[0.6\]
B) \[1\]
C) \[0.8\]
D) None

Hint:
Use the formula for calculating the conditional probability of two given events which is $P(E \cup F) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)$ and substitute the values of given probability of events. The union of two or more sets is the set that contains all the elements of the both or more sets. Union is denoted by the symbol \[ \cup \].
The general probability addition rule for the union of two events states that \[P(A \cup B) = P\left( A \right) + P\left( B \right) - P(A \cap B)\], where \[A \cap B\] is the intersection of the two sets.
The intersection of two or more sets is the set of elements that are common to both or every set. The symbol \[ \cap \]is used to denote the intersection.

Complete step by step solution:
We have been given two events E and F in a sample space such that$P\left( E \right) = 0.5,P(E \cup F) = 0.8,P\left( {E \cap F} \right) = 0.3$.
We have to find the value of \[P\left( F \right)\].
We will use the formula for probability which says that $P(E \cup F) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)$.
By substituting the values provided in the question $P\left( E \right) = 0.5,P(E \cup F) = 0.8,P\left( {E \cap F} \right) = 0.3$ in the above formula, we get
$P(E \cup F) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)$
$ \Rightarrow 0.8 = 0.5 + P\left( F \right) - 0.3$
\[ \Rightarrow P\left( F \right) = 0.8 - 0.5 + 0.3 = 0.6\]
Thus, we have \[P\left( F \right) = 0.6\].

So, option (A) is the correct answer.

Note:
The addition rule is reduced if the sets are disjoint: \[P(A \cup B) = P\left( A \right) + P\left( B \right)\]. The same can be extended to more than two sets if they are all disjoint: \[P(A \cup B \cup C) = P\left( A \right) + P\left( B \right) + P\left( C \right)\].
When events are independent, then we can use the multiplication rule for independent events, which expresses as \[P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)\].
Probability of any event is defined by how likely an event will have to occur or how likely it is that a proposition holds true. The value of probability of any event always lies in the range $\left[ {0,1} \right]$ where having $0$ probability indicates that the event is impossible to happen, while having probability equal to $1$ indicates that the event will surely happen.
The most important point in probability is that the sum of probability of occurrence of any event and probability of non-occurrence of the same event is always $1$.