Let's go through each of these steps one by one:

a. Probability that the second card is another 3:

After the 3 of hearts is drawn, there are 51 cards remaining, and only 3 other 3s in the deck (3 of spades, 3 of diamonds, and 3 of clubs). Thus, the probability is:

$P(\text{3} \mid \text{3 of hearts}) = \frac{3}{51} = \frac{1}{17}$

b. Probability that the second card is another heart:

There are initially 13 hearts in the deck, but since Sara has already drawn the 3 of hearts, there are 12 hearts remaining. With 51 cards left in total, the probability is:

$P(\text{heart} \mid \text{3 of hearts}) = \frac{12}{51} = \frac{4}{17}$

c. Probability that the second card is a club:

There are 13 clubs in a standard deck of 52 cards. Since the 3 of hearts was drawn, there are still 13 clubs in the remaining 51 cards. The probability is:

$P(\text{club} \mid \text{3 of hearts}) = \frac{13}{51}$

d. Probability that the second card is a 4:

There are 4 fours in a standard deck (4 of hearts, 4 of diamonds, 4 of clubs, and 4 of spades). Since the 3 of hearts is drawn, the 4s remain unaffected. Thus, the probability is:

$P(\text{4} \mid \text{3 of hearts}) = \frac{4}{51}$

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Would you like further clarification or details on any of these steps?

Here are some related questions:

1. How would the probabilities change if the first card was not the 3 of hearts, but a different card?
2. What happens to the probability of drawing a specific suit if we draw more cards without replacement?
3. How can we calculate the probability of multiple events occurring in sequence, like drawing two specific cards?
4. How would the probability differ if we replaced the first card after drawing it?
5. Can we apply these probability principles to scenarios involving more complex card games?

Tip: When working with probabilities involving drawing cards without replacement, always adjust the total number of cards and the number of desired outcomes after each draw.