A bag contains 44 U.S. quarters and six Canadian quarters. (The coins are identical in size.) If four quarters are randomly picked from the bag, what is the probability of getting at least one Canadian quarter? (Round your answer to one decimal place.)

To find the probability of getting at least one Canadian quarter when four quarters are randomly picked from the bag, we can follow these steps:

### Step 1: Calculate the total number of quarters.
- Total quarters = 44 U.S. quarters + 6 Canadian quarters = 50 quarters.

### Step 2: Calculate the probability of getting zero Canadian quarters.
This can be done by first calculating the probability of picking only U.S. quarters in all four picks.

- The probability of picking a U.S. quarter on the first draw:
  \[
  \frac{44}{50}
  \]

- The probability of picking a U.S. quarter on the second draw (assuming the first was U.S.):
  \[
  \frac{43}{49}
  \]

- The probability of picking a U.S. quarter on the third draw:
  \[
  \frac{42}{48}
  \]

- The probability of picking a U.S. quarter on the fourth draw:
  \[
  \frac{41}{47}
  \]

So, the probability of picking four U.S. quarters (and hence zero Canadian quarters) is:
\[
\text{P}(0 \text{ Canadian quarters}) = \frac{44}{50} \times \frac{43}{49} \times \frac{42}{48} \times \frac{41}{47}
\]

### Step 3: Calculate the probability of getting at least one Canadian quarter.
This is the complement of the probability of getting zero Canadian quarters:
\[
\text{P}(\text{at least 1 Canadian quarter}) = 1 - \text{P}(0 \text{ Canadian quarters})
\]

Let's calculate this.

### Step 4: Perform the calculation.

\[
\text{P}(0 \text{ Canadian quarters}) = \frac{44}{50} \times \frac{43}{49} \times \frac{42}{48} \times \frac{41}{47} \approx 0.5097
\]

\[
\text{P}(\text{at least 1 Canadian quarter}) = 1 - 0.5097 \approx 0.4903
\]

### Step 5: Round the answer to one decimal place.

\[
\text{P}(\text{at least 1 Canadian quarter}) \approx 0.5
\]

So, the probability of getting at least one Canadian quarter is **0.5** or **50%**.

Would you like more details, or do you have any questions?

Here are 5 related questions for practice:
1. What is the probability of getting exactly two Canadian quarters?
2. How would the probability change if there were 60 quarters in total with 10 being Canadian?
3. What is the probability of getting at least two Canadian quarters?
4. If only two quarters are drawn, what is the probability of getting at least one Canadian quarter?
5. How does the probability change if the number of Canadian quarters is reduced to 4?

**Tip:** Understanding how to calculate probabilities involving "at least" one event often involves using the complement rule, which can simplify complex calculations.

Calculate the probability of getting at least one Canadian quarter when four quarters are randomly picked from a bag containing 44 U.S. quarters and 6 Canadian quarters. This problem utilizes combinatorics and the complement rule, making it suitable for Grades 9-12.

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