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)
Responses

2/9
Start Fraction 2 over 9 End Fraction

2/3
Start Fraction 2 over 3 End Fraction

7/27
Start Fraction 7 over 27 End Fraction

1/9

Answer 1

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:

1. \(\binom{3}{2} = 3\) (the number of ways to choose 2 successes from 3 trials)
2. \(\left(\frac{1}{3}\right)^2 = \frac{1}{9}\) (the probability of picking a dime twice)
3. \(\left(\frac{2}{3}\right)^{1} = \frac{2}{3}\) (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:

\[
\boxed{\frac{2}{9}}
\]