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Sagot :
To find the probability that the coin selected is either a dime or a Canadian coin, let's break the problem into a few steps:
1. Determine the total number of coins in the pocket:
- Number of nickels: 4
- Number of dimes: 7
- Therefore, the total number of coins is [tex]\(4 + 7 = 11\)[/tex].
2. Determine the number of Canadian coins:
- Number of Canadian nickels: 1
- Number of Canadian dimes: 1
- Therefore, the total number of Canadian coins is [tex]\(1 + 1 = 2\)[/tex].
3. Identify the favorable outcomes (i.e., the coin selected is either a dime or a Canadian coin):
- Number of dimes: 7
- Number of Canadian coins: 2
- Since one of the Canadian coins is also counted in the dimes, it should not be counted twice. Total favorable outcomes is the number of dimes plus the remaining (non-dime) Canadian coins:
- Dimes: 7
- Additional Canadian coin (the Canadian nickel that is not a dime): 1
- Therefore, the total number of favorable outcomes is [tex]\(7 + 2 = 9\)[/tex].
4. Calculate the probability:
- The total number of outcomes is the total number of coins: 11
- The number of favorable outcomes is 9
- Therefore, the probability [tex]\(P\)[/tex] that the coin selected is either a dime or a Canadian coin is given by:
[tex]\[ P(\text{dime or Canadian}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{9}{11} \][/tex]
Thus, the probability that the randomly selected coin is either a dime or a Canadian coin is [tex]\(\frac{9}{11}\)[/tex].
Given the options, [tex]\(\frac{9}{11}\)[/tex] is not directly listed, but it corresponds to approximately [tex]\(0.8181818181818182\)[/tex], which is the detailed value from the problem statement.
Therefore, the correct answer aligns most closely with the steps provided and the result of the calculation: [tex]\( \boxed{\frac{9}{11}} \)[/tex]. However, since the answer choices do not list [tex]\(\frac{9}{11}\)[/tex], but rather approximate fractions, the exact equivalent isn't offered. Therefore, we recognize the probability without a match in the provided answer choices.
1. Determine the total number of coins in the pocket:
- Number of nickels: 4
- Number of dimes: 7
- Therefore, the total number of coins is [tex]\(4 + 7 = 11\)[/tex].
2. Determine the number of Canadian coins:
- Number of Canadian nickels: 1
- Number of Canadian dimes: 1
- Therefore, the total number of Canadian coins is [tex]\(1 + 1 = 2\)[/tex].
3. Identify the favorable outcomes (i.e., the coin selected is either a dime or a Canadian coin):
- Number of dimes: 7
- Number of Canadian coins: 2
- Since one of the Canadian coins is also counted in the dimes, it should not be counted twice. Total favorable outcomes is the number of dimes plus the remaining (non-dime) Canadian coins:
- Dimes: 7
- Additional Canadian coin (the Canadian nickel that is not a dime): 1
- Therefore, the total number of favorable outcomes is [tex]\(7 + 2 = 9\)[/tex].
4. Calculate the probability:
- The total number of outcomes is the total number of coins: 11
- The number of favorable outcomes is 9
- Therefore, the probability [tex]\(P\)[/tex] that the coin selected is either a dime or a Canadian coin is given by:
[tex]\[ P(\text{dime or Canadian}) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}} = \frac{9}{11} \][/tex]
Thus, the probability that the randomly selected coin is either a dime or a Canadian coin is [tex]\(\frac{9}{11}\)[/tex].
Given the options, [tex]\(\frac{9}{11}\)[/tex] is not directly listed, but it corresponds to approximately [tex]\(0.8181818181818182\)[/tex], which is the detailed value from the problem statement.
Therefore, the correct answer aligns most closely with the steps provided and the result of the calculation: [tex]\( \boxed{\frac{9}{11}} \)[/tex]. However, since the answer choices do not list [tex]\(\frac{9}{11}\)[/tex], but rather approximate fractions, the exact equivalent isn't offered. Therefore, we recognize the probability without a match in the provided answer choices.
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