1. **Stating the problem:** We want to find the probability of drawing a club on the first draw and a Jack on the second draw, without replacement, from a standard 52-card deck. 2. **Step 1 - Probability of drawing a club on first draw:** There are 13 clubs in the deck of 52 cards, so: $$P(\text{club first})=\frac{13}{52}=\frac{1}{4}$$ 3. **Step 2 - Probability of drawing a Jack on second draw given club first:** There are 4 Jacks in the deck. We consider two cases: - Case 1: The first card drawn was the Jack of clubs. Since we have taken one club which is also a Jack, only 3 Jacks remain out of 51 cards. - Case 2: The first card drawn was a club but not a Jack; all 4 Jacks remain out of 51 cards. The probability of drawing the Jack of clubs first is $\frac{1}{52}$ (since one card is the Jack of clubs). The probability of drawing a club but not Jack first is $\frac{12}{52}$. 4. **Step 3 - Combine probabilities:** The total probability is \begin{align*} P &= P(\text{Jack of clubs first}) \times P(\text{Jack second} | \text{Jack of clubs first}) \\ &\quad + P(\text{club but not Jack first}) \times P(\text{Jack second} | \text{club but not Jack first}) \\ &= \frac{1}{52} \times \frac{3}{51} + \frac{12}{52} \times \frac{4}{51} \\ &= \frac{1 \times 3 + 12 \times 4}{52 \times 51} = \frac{3 + 48}{2652} = \frac{51}{2652} = \frac{1}{52} \end{align*} 5. **Final answer:** The probability of drawing a club then a Jack (without replacement) is $$\boxed{\frac{1}{52}}.$$