To solve this problem, we can use the binomial probability formula since the draws are independent, and the coin is replaced after each draw.
The probability of picking a dime (\(p\)) is \( \frac{1}{3} \) since there are three coins (nickel, penny, and dime). The probability of not picking a dime (\(q\)) is \(1 - p = \frac{2}{3}\).
We want to find the probability of picking a dime exactly 2 times in 3 trials. The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
where:
- \(n\) = total number of trials (3)
- \(k\) = number of successful outcomes (2)
- \(p\) = probability of success (picking a dime)
- \(q\) = probability of failure (not picking a dime)
Calculating:
- \(\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3\)
- \(p^k = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)
- \(q^{n-k} = \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}\)
Now plug these values into the formula:
\[ P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9} \]
Thus, the probability that a dime is picked exactly twice is
\[ \frac{2}{9}. \]
The correct response is:
Start Fraction 2 over 9 End Fraction.