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Question 1
a. For the first card, there are 13 spades out of 52 cards to choose from. For the second card, there are 12 spades left out of 51 remaining cards to choose from. This makes the probability [tex]\frac{13}{52} \cdot \frac{12}{51}=\frac{1}{17}[/tex].
b. There are four possible suits, each of which has the same probability of having two cards drawn from it. Therefore, the answer is [tex]4 \cdot \frac{1}{17}=\frac{4}{17}[/tex].
c. For the first card, there are 4 kings out of 52 cards to choose from. For the second card, there are 3 kings left out of 51 remaining cards, making the probability [tex]\frac{4}{52} \cdot \frac{3}{51}=\frac{1}{221}[/tex].
Question 2
a. There is a [tex]\frac{4}{52}[/tex] probability of the first card being a jack, a [tex]\frac{3}{51}[/tex] probability of the second card being a jack, and a [tex]\frac{2}{50}[/tex] probability of the third card being a jack. This makes the answer [tex]\frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50}=\frac{1}{5525}[/tex].
b. The probability of the first card being an ace is [tex]\frac{4}{52}[/tex]. The probability of the second card being a king is [tex]\frac{4}{51}[/tex]. The probability of the third card being a queen is [tex]\frac{4}{50}[/tex]. This makes the answer [tex]\frac{4}{52} \cdot \frac{4}{51} \cdot \frac{4}{50}=\frac{8}{16575}[/tex].
c. The probability of the first card being a club is [tex]\frac{13}{52}[/tex]. The probability of the second card being a spade is [tex]\frac{13}{51}[/tex]. The probability of the third card being a heart is [tex]\frac{13}{50}[/tex]. This makes the answer [tex]\frac{13}{52} \cdot \frac{13}{51} \cdot \frac{13}{50}=\frac{169}{10200}[/tex].
d. The probability of the first card being a club is [tex]\frac{13}{52}[/tex]. The probability of the second card being a club is [tex]\frac{12}{51}[/tex]. The probability of the third card being a club is [tex]\frac{11}{50}[/tex]. This makes the answer [tex]\frac{13}{52} \cdot \frac{12}{51} \cdot \frac{11}{50}=\frac{11}{850}[/tex].
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