To solve this problem, we can use the binomial probability formula. The probability of picking a dime exactly twice in three picks, when the coin is replaced after each draw, can be calculated using the following parameters:
- Let \( n = 3 \) (the number of trials).
- Let \( k = 2 \) (the number of successful trials, which is picking a dime).
- The probability of success (picking a dime) in each trial \( p = \frac{1}{3} \) (as there is one dime out of three coins).
- The probability of failure (not picking a dime) in each trial \( q = 1 - p = \frac{2}{3} \).
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Where \( \binom{n}{k} \) is the binomial coefficient which represents the number of ways to choose \( k \) successes in \( n \) trials.
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Calculate the binomial coefficient: \[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3 \]
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Calculate \( p^k \): \[ p^k = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]
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Calculate \( q^{n-k} \): \[ q^{n-k} = \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 = \frac{2}{3} \]
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Now, plug these values into the binomial probability formula: \[ P(X = 2) = \binom{3}{2} p^2 q^1 = 3 \times \frac{1}{9} \times \frac{2}{3} \]
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Simplify the expression: \[ P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9} \]
So, the probability that a dime is picked exactly twice is \( \frac{2}{9} \).
The correct response is: Start Fraction 2 over 9 End Fraction.