Statistics and Probability - Mutually exclusive events

A box contains $4$ red, $3$ yellow, and $5$ green marbles. If one marble is selected at random, what is the probability that the marble is either red or yellow?

The total number of marbles is $4 + 3 + 5 = 12$. Let $R$ be the event of picking a red marble: $P(R) = \frac{4}{12} = \frac{1}{3}$. Let $Y$ be the event of picking a yellow marble: $P(Y) = \frac{3}{12} = \frac{1}{4}$. Since the events are mutually exclusive (a marble cannot be both red and yellow), we apply the addition rule: $P(R \cup Y) = P(R) + P(Y) = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}$.

Since only one marble is drawn, the outcomes 'Red' and 'Yellow' cannot happen at the same time. Therefore, we simply add the probabilities of the individual events.

In a deck of $52$ playing cards, are the events 'Drawing a King' and 'Drawing a Queen' mutually exclusive? Find the probability of drawing either a King or a Queen in a single draw.

Yes, they are mutually exclusive because a card cannot be both a King and a Queen. $P(\text{King}) = \frac{4}{52} = \frac{1}{13}$ $P(\text{Queen}) = \frac{4}{52} = \frac{1}{13}$ $P(\text{King or Queen}) = P(\text{King}) + P(\text{Queen}) = \frac{1}{13} + \frac{1}{13} = \frac{2}{13}$.

To find the probability of 'or' for mutually exclusive events, we use the specific addition rule $P(A \cup B) = P(A) + P(B)$.

Events $A$ and $B$ are mutually exclusive. Given $P(A) = 0.35$ and $P(B) = 0.45$, calculate $P(A \text{ or } B)$ and $P(A \text{ and } B)$.

$P(A \cup B) = 0.35 + 0.45 = 0.80$ $P(A \cap B) = 0$

For mutually exclusive events, the probability of both occurring ($A$ and $B$) is always $0$. The probability of either occurring ($A$ or $B$) is the sum of their probabilities.