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

23
Start Fraction 2 over 3 End Fraction

29
Start Fraction 2 over 9 End Fraction

19
Start Fraction 1 over 9 End Fraction

727
Start Fraction 7 over 27 End Fraction

Answer 1

To find the probability of picking a dime exactly twice in three draws with replacement, we can use the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

where:
- \( n \) is the total number of draws (3 in this case),
- \( k \) is the number of successful draws (2 for picking a dime),
- \( p \) is the probability of picking a dime,
- \( (1-p) \) is the probability of not picking a dime (which would be picking either a nickel or a penny).

Since there are a total of 3 coins (a nickel, a penny, and a dime), the probability of picking a dime (\( p \)) is:

\[
p = \frac{1}{3}
\]

The probability of not picking a dime is:

\[
1 - p = \frac{2}{3}
\]

Now, substituting the known values (\( n = 3 \), \( k = 2 \), \( p = \frac{1}{3} \)) into the binomial formula:

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

First, calculate \( \binom{3}{2} \):

\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 1} = 3
\]

Next, calculate \( \left(\frac{1}{3}\right)^2 \):

\[
\left(\frac{1}{3}\right)^2 = \frac{1}{9}
\]

And \( \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 \):

\[
\left(\frac{2}{3}\right) = \frac{2}{3}
\]

Putting it all together:

\[
P(X = 2) = 3 \cdot \frac{1}{9} \cdot \frac{2}{3}
\]

Calculating this gives:

\[
P(X = 2) = 3 \cdot \frac{2}{27} = \frac{6}{27} = \frac{2}{9}
\]

Thus, the probability that a dime is picked exactly twice is:

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