Probability - Algebra of events

In an experiment of rolling a fair die, let $A$ be the event of getting an even number and $B$ be the event of getting a multiple of $3$. Find $A \cup B$ and $A \cap B$.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Event $A = \{2, 4, 6\}$ and Event $B = \{3, 6\}$. To find $A \cup B$, we combine all elements: $A \cup B = \{2, 3, 4, 6\}$. To find $A \cap B$, we look for common elements: $A \cap B = \{6\}$.

Union represents 'either or both', while intersection represents 'only common elements'.

If $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$, calculate the probability that neither $A$ nor $B$ occurs.

First, find $P(A \cup B)$: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $P(A \cup B) = 0.6 + 0.5 - 0.2 = 0.9$

The event 'neither $A$ nor $B$' is $(A \cup B)'$. $P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.9 = 0.1$.

The probability of the complement of the union gives the probability that none of the events occur.

Calculate the total number of favorable outcomes for $n(A \cup B)$ if $n(A) = 45$, $n(B) = 32$, and $n(A \cap B) = 12$.

Using the principle of inclusion-exclusion: $\begin{array}{r} 45 \\ + 32 \\ \hline 77 \\ - 12 \\ \hline 65 \end{array}$ $n(A \cup B) = 65$.

We add the elements of both sets and subtract the intersection to avoid double-counting.