Probability - Mutually exclusive events

A die is thrown. Let event $A$ be 'getting an odd number' and event $B$ be 'getting an even number'. Are these events mutually exclusive? Find $P(A \cup B)$.

The sample space is $S = \{1, 2, 3, 4, 5, 6\}$. Event $A = \{1, 3, 5\}$, so $P(A) = \frac{3}{6} = \frac{1}{2}$. Event $B = \{2, 4, 6\}$, so $P(B) = \frac{3}{6} = \frac{1}{2}$. Since $A \cap B = \phi$, the events are mutually exclusive. $P(A \cup B) = P(A) + P(B)$ $P(A \cup B) = \frac{1}{2} + \frac{1}{2} = 1$

Because a single roll of a die cannot result in a number that is both odd and even, the intersection is empty. Therefore, we simply add the probabilities.

If $P(A) = 0.35$, $P(B) = 0.45$, and $P(A \cup B) = 0.8$, show that $A$ and $B$ are mutually exclusive events.

We use the general addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting the given values: $0.8 = 0.35 + 0.45 - P(A \cap B)$ $0.8 = 0.8 - P(A \cap B)$ $P(A \cap B) = 0.8 - 0.8 = 0$

Since the probability of the intersection $P(A \cap B)$ is calculated to be $0$, the events $A$ and $B$ must be mutually exclusive.

Two cards are drawn from a pack of 52 cards. Let $E_1$ be the event of drawing a red card and $E_2$ be the event of drawing a black card. Find the probability that the card drawn is either red or black.

There are 26 red cards and 26 black cards in a deck. $P(E_1) = \frac{26}{52} = \frac{1}{2}$ $P(E_2) = \frac{26}{52} = \frac{1}{2}$ A card cannot be both red and black, so $E_1 \cap E_2 = \phi$. $P(E_1 \cup E_2) = P(E_1) + P(E_2)$ $P(E_1 \cup E_2) = \frac{1}{2} + \frac{1}{2} = 1$

Since the sets of red cards and black cards are disjoint, we use the addition rule for mutually exclusive events. The result $1$ indicates these events are also exhaustive.