Why do you think the first answer requires independence? If we did it with independence the probability the first is a heart then a a spade would be $\frac 14\cdot \frac 14$ and the probability of a spade then a heart would be $\frac 14\cdot \frac 14$ and the probability would be $\frac 14 \frac 14 + \frac 14 \frac 14 = \frac 18$.
That's not quite right. What does "$S$" mean? Does it mean a specific card is a spade? Or does it mean at least one of two cards is a spade? Or something else. If $P(S) = \frac 14$ this seems to imply you mean a specific card is a spade. But then $P(S\cap H)$ would mean the probability that a specific card is both a heart and a spade and $P(H|S)$ is the probability of a card being a spade given we know it is a heart. (So $0 = P(S\cap H) = P(S)P(H|S) = \frac 14\cdot 0 = 0$.) Or maybe $S$ means a specific card is a spade, and $H$ means the other card is a heart. Then your calculation is correct..... But you figured the probability that a specific card is a spade and the other specific card is a heart.... And that was not the question. The question was that either card is a spade and the other card is a heart. If $S$ at least one card of two is a spade and $H$ is at least one card of two is a heart then to calculate conditional probability would go like this: $P(H\cap S) = P(S)P(H|S)$ is $\frac {13*39 + 39*13+13*13}{52*51}\cdot \frac{ 13*13 + 13*13}{13*39 + 39*13 + 13*13}=$ $\frac {13\cdot 26}{52*51}= \frac {13}{102}$ But that's a ridiculously hard way to do it. Better to either figure there are $2\times 13 \times 13$ (heart, spade) and (spade, heart) pairs where order matters out of $52\times 51$ combos; or there are $13\times 13$ (heart,spade) pairs were order doesn't matter out of ${52\choose 2}$ combos. ========= tl;dr you figured out the probability of a specific card being a spade and the other not being. As order does not matter the probability is one half of that.
Exercise :: Probability - General Questions
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In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
Q: Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4. Answer & Explanation Answer: D) 6 Explanation: |