Probability - Probability of the event 'A or B'

A card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is either a 'Red card' or a 'King'.

Let $A$ be the event that the card is Red, and $B$ be the event that the card is a King. Total outcomes $n(S) = 52$. Number of Red cards $n(A) = 26$, so $P(A) = \frac{26}{52}$. Number of Kings $n(B) = 4$, so $P(B) = \frac{4}{52}$. There are 2 Red Kings (King of Hearts and King of Diamonds), so $n(A \cap B) = 2$ and $P(A \cap B) = \frac{2}{52}$. Using the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $P(A \cup B) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52} = \frac{28}{52} = \frac{7}{13}$

We identify $P(A)$ and $P(B)$, but since there is an overlap (Red Kings), we subtract the intersection to avoid double-counting.

A die is thrown once. Find the probability of getting a number divisible by 2 or 3.

Sample Space $S = \{1, 2, 3, 4, 5, 6\}$, so $n(S) = 6$. Let $E_1$ be the event of getting a number divisible by 2: $E_1 = \{2, 4, 6\} \implies P(E_1) = \frac{3}{6}$. Let $E_2$ be the event of getting a number divisible by 3: $E_2 = \{3, 6\} \implies P(E_2) = \frac{2}{6}$. The intersection $E_1 \cap E_2 = \{6\}$ (numbers divisible by both 2 and 3), so $P(E_1 \cap E_2) = \frac{1}{6}$. Using the addition theorem: $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$ $P(E_1 \cup E_2) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$

To find the probability of '2 or 3', we sum the individual probabilities and subtract the probability of the number that satisfies both conditions (the number 6).

Given $P(A) = 0.5$, $P(B) = 0.35$, and $P(A \cap B) = 0.15$. Find $P(A \text{ or } B)$.

The event '$A$ or $B$' is $A \cup B$. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ Substituting the values: $P(A \cup B) = 0.5 + 0.35 - 0.15$ $P(A \cup B) = 0.85 - 0.15 = 0.70$

Direct application of the Addition Theorem for two events.