What are the probabilities of these card situations?

Your b) part needs some work.
You must consider two cases: an ace is the first card and the first card is not an ace.
 
Hello, dhs316!

Two cards are drawn successively without replacement from a deck of cards.
Compute the probability of drawing:

b) a heart (of ANY face value) on the first draw and an ace on the second draw.
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There are two cases to consider:

. . \(\displaystyle \text{[1] The }A\heartsuit \text{ is drawn on the first draw.}\)

. . \(\displaystyle \text{[2] Any }other\text{ heart is drwn on the first draw.}\)


\(\displaystyle [1]\!:\:p(A\heartsuit\text{, 1st draw}) \,=\,\frac{1}{52} \qquad P(\text{Ace, 2nd draw}) \:=\:\frac{3}{51}\)

. . .\(\displaystyle \text{Hence: }\;P\bigg((A\heartsuit\text{ 1st draw}) \,\wedge\,(\text{Ace, 2nd draw})\bigg) \:=\:\frac{1}{52}\cdot\frac{3}{51} \:=\:\frac{3}{2652}\)


\(\displaystyle [2]\!:\;P(\text{other }\heartsuit\text{. 1st draw}) \:=\:\frac{12}{52}\qquad P(\text{Ace, 2nd draw}) \:=\:\frac{4}{51}\)

. . /\.\(\displaystyle \text{Hence: }\:p\bigg(\text{(other }\heartsuit\text{, 1st draw})\,\wedge\,(\text{Ace, 2nd draw})\bigg) \:=\:\frac{12}{52}\cdot\frac{4}{51} \:=\:\frac{48}{2652}\)


\(\displaystyle \text{Therefore: }\:p((\heartsuit\text{, 1st draw} \,\wedge\,(\text{Ace, 2nd draw})) \;=\;\frac{3}{2652} + \frac{48}{2652} \;=\;\frac{51}{2652} \;=\;\frac{1}{52}\)

 
Thank you! i never realized it was that important to think of all the scenarios
 
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