From a pack of $52$ cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade ?

### Concept & Probability of Non-Mutually Exclusive Events To find the probability of $A$ or $B$ occurring, we must add the probabilities of $A$ and $B$, then subtract the probability of both occurring so we don't double-count the overlap. $$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$$ ### Step-by-Step Solution * **Total Outcomes:** $n(S) = 52$. * **Event A (Ten):** There are $4$ tens. $P(\text{Ten}) = \frac{4}{52}$. * **Event B (Spade):** There are $13$ spades. $P(\text{Spade}) = \frac{13}{52}$. * **Event A and B (Ten of Spades):** There is $1$ ten of spades. $P(\text{Ten} \cap \text{Spade}) = \frac{1}{52}$. * **Calculation:** $P(\text{Ten} \cup \text{Spade}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$. * Simplify the fraction by dividing numerator and denominator by $4$: $\frac{16}{52} = \frac{4}{13}$. ### Exam Strategy & Shortcut Count manually to avoid formula errors: We have $13$ spades. We also have $3$ other tens (clubs, hearts, diamonds). Total favorable cards = $13 + 3 = 16$. The probability is $\frac{16}{52} = \frac{4}{13}$. ### Common Pitfall Forgetting to subtract the intersecting card (the Ten of Spades), which leads to the incorrect probability of $\frac{17}{52}$. ### Final Answer Therefore, the correct answer is **$\frac{4}{13}$**.