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 po int)
Responses

19
Start Fraction 1 over 9 End Fraction

29
Start Fraction 2 over 9 End Fraction

23
Start Fraction 2 over 3 End Fraction

727

Answer 1

To find the probability of picking a dime exactly twice when randomly drawing from the bag three times (with replacement), we can use the binomial probability formula:

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

Where:
- \( n = 3 \) (the number of trials)
- \( k = 2 \) (the number of successes, which in this case is picking a dime)
- \( p \) is the probability of success on an individual trial.

First, we identify the probabilities:
- The total number of coins = 3 (a nickel, a penny, and a dime).
- The probability of picking a dime \( p = \frac{1}{3} \).
- The probability of not picking a dime \( 1-p = \frac{2}{3} \).

Now we substitute the values into the binomial formula:

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

Calculating the binomial coefficient:

\[
\binom{3}{2} = 3
\]

Now substituting all the values:

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

Calculating each term:

\[
= 3 \cdot \frac{1}{9} \cdot \frac{2}{3}
\]
\[
= 3 \cdot \frac{2}{27}
\]
\[
= \frac{6}{27}
\]
\[
= \frac{2}{9}
\]

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

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

The correct answer is:

**Start Fraction 2 over 9 End Fraction**.