Two cards are drawn together from a pack of $52$ cards. The probability that one is a spade and one is a heart, is

### Concept & Combinations for Multiple Independent Sets When selecting items from different distinct groups (like different suits), we multiply the combinations of selecting from each group. $$\text{Favorable Outcomes} = \binom{n_1}{r_1} \times \binom{n_2}{r_2}$$ ### Step-by-Step Solution * **Total Outcomes:** Number of ways to draw $2$ cards from $52$. * $n(S) = \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326$. * **Favorable Outcomes:** We need $1$ spade (out of $13$) and $1$ heart (out of $13$). * Ways to choose $1$ spade = $\binom{13}{1} = 13$. * Ways to choose $1$ heart = $\binom{13}{1} = 13$. * Total favorable ways $n(E) = 13 \times 13 = 169$. * **Calculation:** $P(E) = \frac{n(E)}{n(S)} = \frac{169}{1326}$. * Both numbers are divisible by $13$: $\frac{169 \div 13}{1326 \div 13} = \frac{13}{102}$. ### Exam Strategy & Shortcut Alternatively, use sequential probability and multiply by $2$ (since order doesn't matter: Spade-Heart or Heart-Spade). $P = 2 \times (\text{Prob of Spade 1st} \times \text{Prob of Heart 2nd})$ $P = 2 \times (\frac{13}{52} \times \frac{13}{51}) = 2 \times \frac{1}{4} \times \frac{13}{51} = \frac{1}{2} \times \frac{13}{51} = \frac{13}{102}$. ### Common Pitfall Using sequential probabilities but forgetting to multiply by $2$ for the different possible arrangements (Spade then Heart OR Heart then Spade). ### Final Answer Therefore, the correct answer is **$\frac{13}{102}$**.