 Two cards are drawn without replacement from a standard 52-card deck. What is the probability that the second card drawn is a jack?

To solve this problem, we need to consider the different scenarios that lead to the second card being a jack. We will use conditional probability to account for the first card drawn affecting the outcome of the second draw. 

### Total Possible Outcomes
When drawing two cards from a deck of 52 without replacement, the total number of possible outcomes is:
\[
\binom{52}{2} = \frac{52 \times 51}{2} = 1326
\]

### Favorable Outcomes
To determine the probability that the second card is a jack, we consider the following cases:

1. The first card is a jack.
2. The first card is not a jack.

#### Case 1: The first card is a jack
If the first card is a jack, there are 3 jacks left in a deck of 51 cards.
\[
P(\text{second card is a jack} \mid \text{first card is a jack}) = \frac{3}{51}
\]

The probability that the first card is a jack:
\[
P(\text{first card is a jack}) = \frac{4}{52}
\]

Therefore, the probability for this case is:
\[
P(\text{first card is a jack}) \times P(\text{second card is a jack} \mid \text{first card is a jack}) = \frac{4}{52} \times \frac{3}{51}
\]

#### Case 2: The first card is not a jack
If the first card is not a jack, there are still 4 jacks left in a deck of 51 cards.
\[
P(\text{second card is a jack} \mid \text{first card is not a jack}) = \frac{4}{51}
\]

The probability that the first card is not a jack:
\[
P(\text{first card is not a jack}) = \frac{48}{52}
\]

Therefore, the probability for this case is:
\[
P(\text{first card is not a jack}) \times P(\text{second card is a jack} \mid \text{first card is not a jack}) = \frac{48}{52} \times \frac{4}{51}
\]

### Total Probability
Adding the probabilities of both cases:
\[
\text{Total Probability} = \left(\frac{4}{52} \times \frac{3}{51}\right) + \left(\frac{48}{52} \times \frac{4}{51}\right)
\]

Simplifying the fractions:
\[
\text{Total Probability} = \frac{4}{52} \times \frac{3}{51} + \frac{48}{52} \times \frac{4}{51}
\]
\[
\text{Total Probability} = \frac{1}{13} \times \frac{3}{51} + \frac{12}{13} \times \frac{4}{51}
\]
\[
\text{Total Probability} = \frac{3}{663} + \frac{48}{663}
\]
\[
\text{Total Probability} = \frac{3 + 48}{663}
\]
\[
\text{Total Probability} = \frac{51}{663} = \frac{1}{13}
\]

So, the probability that the second card drawn is a jack is \(\frac{1}{13}\).

Would you like any further details or explanations?

Here are 5 numbered questions you can ask:
1. How does drawing without replacement affect probabilities?
2. Can you explain the concept of conditional probability?
3. What are the basic rules of probability?
4. How do you calculate probabilities for card draws in general?
5. Can you provide another example involving probability with a standard deck of cards?

Tip: When solving probability problems, clearly define each step and consider all possible scenarios to ensure accurate calculations.

Explore the probability of drawing a jack as the second card from a standard 52-card deck. Learn about conditional probability and calculate the total probability using different scenarios. Suitable for high school students studying probability.

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