A random variable \[X\] has a probability distribution.

If the event \[E = \left\{ \text{X is prime number} \right\}\] and \[F = \left\{ {X < 4} \right\}\]. Then what is the value of \[P\left( {E \cup F} \right)\]?
A. \[0.77\]
B. \[0.87\]
C. \[0.35\]
D. \[0.50\]

Hint: First, use the definition of the events and find the elements of the sets. Then use the given probability distribution and find the probabilities of both events. After that, find the probability of both events occurring together. In the end, use the addition rule of probability to get the required answer.

Formula Used:
Addition rule: \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]

Complete step by step solution:
The given probability distribution of a random variable \[X\] is

And the events are \[E = \left\{ \text{X is prime number} \right\}\] and \[F = \left\{ {X < 4} \right\}\].

Let’s find out the elements of the events.
We have, \[X = \left\{ {1,2,3,4,5,6,7,8} \right\}\].
Then,
\[E = \left\{ \text{X is prime number} \right\} = \left\{ {2,3,5,7} \right\}\]
\[F = \left\{ {X < 4} \right\} = \left\{ {1,2,3} \right\}\]

Now calculate the probabilities of the both events.
\[P\left( E \right) = P\left( 2 \right) + P\left( 3 \right) + P\left( 5 \right) + P\left( 7 \right)\]
Substitute the values in the above equation.
\[P\left( E \right) = \left( {0.23} \right) + \left( {0.12} \right) + \left( {0.20} \right) + P\left( {0.07} \right)\]
\[ \Rightarrow \]\[P\left( E \right) = 0.62\]

\[P\left( F \right) = P\left( 1 \right) + P\left( 2 \right) + P\left( 3 \right)\]
Substitute the values in the above equation.
\[P\left( F \right) = \left( {0.15} \right) + \left( {0.23} \right) + \left( {0.12} \right)\]
\[ \Rightarrow \]\[P\left( F \right) = 0.50\]

Let’s calculate the probability of both events occurring together.
\[\left( {E \cap F} \right) = \left( \text{X should be prime number and less than 4} \right)\]
Then, the probability of the above event is,
\[P\left( {E \cap F} \right) = P\left( 2 \right) + P\left( 3 \right)\]
Substitute the values in the above equation.
\[P\left( {E \cap F} \right) = \left( {0.23} \right) + \left( {0.12} \right)\]
\[ \Rightarrow \]\[P\left( {E \cap F} \right) = 0.35\]

Now apply the addition rule of probability to calculate the probability of \[\left( {E \cup F} \right)\].
\[P\left( {E \cup F} \right) = P\left( E \right) + P\left( F \right) - P\left( {E \cap F} \right)\]
Substitute the values in the above equation.
\[P\left( {E \cup F} \right) = 0.62 + 0.50 - 0.35\]
\[ \Rightarrow \]\[P\left( {E \cup F} \right) = 0.77\]

Hence the correct option is A.

Note: Students often get confused with the addition rule of probability.
The probability of two mutually exclusive events: \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)\]
The probability of two non-mutually exclusive events: \[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\]