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The cards are drawn from a well shuffled deck of 52 cards one after the other without replacement. The probability of first card being a spade and the second a black king is ?

Here, is my approach, Upon first draw we got a black spade king $$P(\text{first card is spade}) = \frac{13}{52}$$ $$ P(\text{second black king}) = \frac{1}{51} $$

Upon first draw we don't get a black spade king $$P(\text{first card is spade}) = \frac{13}{52}$$ $$ P(\text{second black king}) = \frac{2}{51} $$ Now, the total probability, $$P= \frac{13}{52} \frac{1}{51} + \frac{13}{52} \frac{2}{51} = \frac{39}{2652}$$

But the actual answer is $\frac{25}{2652}$

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    For the first case, the probability that the first card is $K\spadesuit$ is $\frac 1{52}$. For the second, the probability that the first card is some spade other than $K\spadesuit$ is $\frac {12}{52}$. – 2017-01-20

2 Answers 2

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Upon first draw we got a black spade king $$P(\text{first card is $K \spadesuit$}) = \frac{1}{52}$$ $$ P(\text{second black king i.e $K\clubsuit$}) = \frac{1}{51} $$

Upon first draw we don't get a black spade king $$P(\text{first card is $\spadesuit$ but not $K\spadesuit$}) = \frac{12}{52}$$ $$ P(\text{second black $K$}) = \frac{2}{51} $$ Now, the total probability, $$P= \frac{1}{52} \frac{1}{51} + \frac{12}{52} \frac{2}{51} = \frac{25}{2652}$$

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As one card is common in both.

So probability = $\frac{13}{52} \cdot \frac{2}{51} -\frac{1}{52 \cdot 51}$

= $\frac{26}{2652} -\frac{1}{2652}$

= $\frac{25}{2652}$