Question

A nickel, a penny, and a dime are placed inside a bag. One coin is randomly picked three times. If the coin is replaced after each draw, what is the probability that a dime is picked exactly twice? (1 point)
2/3
7/27
1/9
2/9

Answer 1

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.

1. **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).

2. **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).

3. **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:
\[
P(X = 2) = \binom{3}{2} \left(\frac{1}{3}\right)^{2} \left(\frac{2}{3}\right)^{1}
\]

4. **Calculate the binomial coefficient:**
\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3}{1} = 3
\]

5. **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 \(\frac{2}{9}\).