A bag contains 44 U.S. quarters and six 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 find the probability of getting at least one Canadian quarter when seven quarters are randomly picked from the bag, we first need to determine the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes:
The total number of quarters in the bag is 44 + 6 = 50. Therefore, the total number of possible outcomes is the number of ways to choose 7 quarters from 50, which can be calculated using the combination formula: C(50, 7) = 50! / (7! * (50-7)!).
Favorable outcomes:
Since there are 6 Canadian quarters, we need to calculate the number of ways to choose 1 to 7 Canadian quarters from the available 6: C(6, 1) + C(6, 2) + C(6, 3) + C(6, 4) + C(6, 5) + C(6, 6) + C(6, 7).
Therefore, the probability of getting at least one Canadian quarter can be calculated as the number of favorable outcomes divided by the total number of possible outcomes:
Probability = (C(6, 1) + C(6, 2) + C(6, 3) + C(6, 4) + C(6, 5) + C(6, 6) + C(6, 7)) / (C(50, 7))
Let's compute this:
Probability = ((6! / (1! * (6-1)!)) + (6! / (2! * (6-2)!)) + (6! / (3! * (6-3)!)) + (6! / (4! * (6-4)!)) + (6! / (5! * (6-5)!)) + (6! / (6! * (6-6)!)) + (6! / (7! * (6-7)!))) / ((50!)/ (7! * (50-7)!))
Simplifying and calculating:
Probability = (6 + 15 + 20 + 15 + 6 + 1 + 0) / 10400600
Probability = 0.001443
Rounded to one decimal place, the probability of getting at least one Canadian quarter is approximately 0.0.