To calculate the probability of picking a quarter first and then a nickel without replacement, let's break this into steps.

Step 1: Total Coins

The total number of coins is:

$4 \ (\text{pennies}) + 2 \ (\text{nickels}) + 2 \ (\text{dimes}) + 4 \ (\text{quarters}) = 12$

Step 2: Probability of Picking a Quarter First

The probability of picking a quarter first is:

$P(\text{quarter first}) = \frac{\text{number of quarters}}{\text{total number of coins}} = \frac{4}{12} = \frac{1}{3}$

Step 3: Probability of Picking a Nickel Second

After picking the quarter, only 11 coins remain, including 2 nickels. The probability of picking a nickel next is:

$P(\text{nickel second | quarter first}) = \frac{\text{number of nickels}}{\text{remaining coins}} = \frac{2}{11}$

Step 4: Combined Probability

Using the multiplication rule for dependent events:

$P(\text{quarter, then nickel}) = P(\text{quarter first}) \cdot P(\text{nickel second | quarter first})$

$P(\text{quarter, then nickel}) = \frac{1}{3} \cdot \frac{2}{11} = \frac{2}{33}$

Final Answer:

$P(\text{quarter, then nickel}) = \frac{2}{33}$

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Would you like a detailed explanation or clarification about this process?
Here are some related questions to consider:

1. What is the probability of picking two quarters consecutively?
2. How does the calculation change if replacement is allowed?
3. What is the probability of picking a nickel first and then a quarter?
4. How would the total probability of picking a quarter or nickel be calculated?
5. What if all coins had the same value? How would probabilities change?

Tip: Always double-check whether the scenario involves replacement or not, as it significantly affects probabilities.