A bag contains 40 U.S. quarters and ten Canadian quarters. (The coins are identical in size.) If seven 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 solve this problem, we need to calculate the probability of drawing at least one Canadian quarter when picking 7 quarters randomly from a bag containing 40 U.S. quarters and 10 Canadian quarters.

### Step-by-Step Solution

1. **Total number of quarters in the bag**:
   \[
   40 \text{ (U.S. quarters)} + 10 \text{ (Canadian quarters)} = 50 \text{ quarters}
   \]

2. **Total number of ways to pick 7 quarters from 50**:
   This can be computed using the binomial coefficient:
   \[
   \binom{50}{7} = \frac{50!}{7!(50-7)!}
   \]

3. **Calculate the probability of getting no Canadian quarters (all quarters picked are U.S. quarters)**:
   - The number of ways to pick 7 quarters from 40 U.S. quarters is:
     \[
     \binom{40}{7} = \frac{40!}{7!(40-7)!}
     \]
   - The probability of picking 7 U.S. quarters out of 7:
     \[
     P(\text{no Canadian quarters}) = \frac{\binom{40}{7}}{\binom{50}{7}}
     \]

4. **Calculate the probability of getting at least one Canadian quarter**:
   - This is the complement of the probability of getting no Canadian quarters.
   \[
   P(\text{at least one Canadian quarter}) = 1 - P(\text{no Canadian quarters})
   \]

Now, let's calculate these values:

\[
\binom{40}{7} = \frac{40!}{7!(33)!} = 18,643,560
\]
\[
\binom{50}{7} = \frac{50!}{7!(43)!} = 99,884,400
\]

So, the probability of getting no Canadian quarters is:
\[
P(\text{no Canadian quarters}) = \frac{18,643,560}{99,884,400} \approx 0.1867
\]

Therefore, the probability of getting at least one Canadian quarter is:
\[
P(\text{at least one Canadian quarter}) = 1 - 0.1867 = 0.8133
\]

### Final Answer

The probability of getting at least one Canadian quarter when 7 quarters are randomly picked from the bag is **0.8133**, which rounds to **0.8** (to one decimal place).

**Answer:** 0.8

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

Here are some related questions for further exploration:
1. What is the probability of drawing exactly two Canadian quarters?
2. How does the probability change if there are 15 Canadian quarters and 35 U.S. quarters in the bag?
3. What is the probability of drawing all Canadian quarters if the number of draws is increased to 10?
4. If two quarters are drawn without replacement, what is the probability that both are Canadian?
5. How would the answer change if the quarters were drawn with replacement?

**Tip:** When solving probability problems, it is often helpful to think about complementary events. Calculating the probability of the complement can sometimes simplify the problem.

Explore the probability of drawing at least one Canadian quarter when seven quarters are randomly picked from a bag containing 40 U.S. quarters and 10 Canadian quarters. Step-by-step solution using combinatorial methods and probability calculations.

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