A group of learners is given the following event sets:

\[\begin{array}{|l|c|c|c|c|}
\hline
\text{Event Set } A & 1 & 2 & 5 & 6 \\
\hline
\end{array}\]
\[\begin{array}{|l|c|}
\hline
\text{Event Set } B & 3 \\
\hline
\end{array}\]

\[\begin{array}{|l|c|}
\hline
\text{Event Set } A \cap B & \text{empty} \\
\hline
\end{array}\]

The sample space can be described as \(\{ n:n \text{ } \epsilon \text{ } \mathbb{Z}, \text{ } 1 \leq n \leq 6 \}\).

They are asked to calculate the value of \(P(A \cup B)\). They get stuck, and you offer to calculate it for them. Give your answer as a decimal number, rounded to two decimal places.

\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]

Identify variables needed:
\begin{align*}
P(A) & = \frac{n(A)}{n(S)} = \frac{4}{6} = \text{0,67} \\
P(B) & = \frac{n(B)}{n(S)} = \frac{1}{6} = \text{0,17} \\
P(A \cap B) & = \frac{n(A \cap B)}{n(S)} = \frac{0}{6} = \text{0}
\end{align*}

Calculate \(P(A \cup B)\):
\begin{align*}
P (A \cup B) & = P(B) + P(A) - P(A \cap B) \\
& = (\text{0,17}) + (\text{0,67}) - (\text{0}) \\
& = \text{0,83}
\end{align*}

Therefore, the value of \(P(A \cup B)\) is \(\text{0,83}\).

A group of learners is given the following event sets:

\[\begin{array}{|l|c|c|c|}
\hline
\text{Event Set } A & 1 & 2 & 6 \\
\hline
\end{array}\]
\[\begin{array}{|l|c|c|}
\hline
\text{Event Set } B & 1 & 5 \\
\hline
\end{array}\]
\[\begin{array}{|l|c|c|c|c|}
\hline
\text{Event Set } A \cup B & 1 & 2 & 5 & 6 \\
\hline
\end{array}\]

The sample space can be described as \(\{ n:n \text{ } \epsilon \text{ } \mathbb{Z}, \text{ } 1 \leq n \leq 6 \}\).

They are asked to calculate the value of \(P(A \cap B)\). They get stuck, and you offer to calculate it for them. Give your answer as a decimal number, rounded to two decimal value.

\[P(A \cup B) = P(A) + P(B) - P(A \cup B)\]

Make \(P(A \cap B)\) the subject, and we get:
\begin{align*}
P (A \cap B) & = P(B) + P(A) - P(A \cup B)
\end{align*}

Identify variables needed:
\begin{align*}
P(A) & = \frac{n(A)}{n(S)} = \frac{3}{6} = \text{0,5} \\
P(B) & = \frac{n(B)}{n(S)} = \frac{2}{6} = \text{0,33} \\
P(A \cup B) & = \frac{n(A \cup B)}{n(S)} = \frac{4}{6} = \text{0,67}
\end{align*}

Calculate \(P(A \cap B)\):
\begin{align*}
P (A \cap B) & = P(B) + P(A) - P(A \cup B) \\
& = (\text{0,33}) + (\text{0,5}) - (\text{0,67}) \\
& = \text{0,17}
\end{align*}

Therefore, the value of \(P(A \cap B)\) is \(\text{0,17}\).