To find the probability of picking a dime exactly twice when drawing from a bag containing a nickel, a penny, and a dime, we can use the binomial probability formula.
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Identify the probability of each event:
- The probability of picking a dime (success) is \( P(D) = \frac{1}{3} \).
- The probability of not picking a dime (failure) is \( P(N) = \frac{2}{3} \) (picking either a nickel or a penny).
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Define the parameters for the binomial distribution:
- Number of trials (n) = 3 (since we draw three times),
- Number of successes (k) = 2 (we want to pick a dime exactly twice).
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Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} (p)^k (1-p)^{n-k} \] where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \),
- \( p \) is the probability of success,
- \( n \) is the number of trials,
- \( k \) is the number of successes.
Substituting the values we have: \[ P(X = 2) = \binom{3}{2} \left(\frac{1}{3}\right)^{2} \left(\frac{2}{3}\right)^{1} \]
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Calculate the binomial coefficient: \[ \binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3}{1} = 3 \]
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Calculate the probability: \[ P(X = 2) = 3 \left(\frac{1}{3}\right)^{2} \left(\frac{2}{3}\right)^{1} \] \[ = 3 \cdot \frac{1}{9} \cdot \frac{2}{3} \] \[ = 3 \cdot \frac{2}{27} = \frac{6}{27} = \frac{2}{9} \]
Thus, the probability that a dime is picked exactly twice is \(\frac{2}{9}\).
The correct answer is \(\frac{2}{9}\).