To solve the problem, we start by identifying the total number of coins and their respective probabilities. The bag contains three coins: a nickel (N), a penny (P), and a dime (D). Therefore, the probability of choosing each type of coin is:
- P(N) = 1/3
- P(P) = 1/3
- P(D) = 1/3
We need to calculate the probability of picking a dime exactly 2 times in 3 draws. This situation follows a binomial distribution where:
- n (number of trials) = 3
- k (number of successes, picking a dime) = 2
- p (probability of success, picking a dime) = 1/3
- q (probability of failure, not picking a dime) = 2/3 (since there are two other coins)
The probability of getting exactly k successes in n trials is given by the formula:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Substituting the values:
\[ P(X = 2) = \binom{3}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^{3-2} \]
Calculating the components:
- \(\binom{3}{2} = 3\) (the number of ways to choose 2 successes from 3 trials)
- \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\) (the probability of picking a dime twice)
- \(\left(\frac{2}{3}\right)^{1} = \frac{2}{3}\) (the probability of not picking a dime once)
Putting it all together:
\[ P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9} \]
Therefore, the probability that a dime is picked exactly twice is:
\[ \boxed{\frac{2}{9}} \]