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 727 Start Fraction 7 over 27 End Fraction 29 Start Fraction 2 over 9 End Fraction 19
Answers
GPT-4o mini
To solve this problem, we can use the binomial probability formula. The probability of picking a dime exactly twice in three picks, when the coin is replaced after each draw, can be calculated using the following parameters:
- Let \( n = 3 \) (the number of trials).
- Let \( k = 2 \) (the number of successful trials, which is picking a dime).
- The probability of success (picking a dime) in each trial \( p = \frac{1}{3} \) (as there is one dime out of three coins).
- The probability of failure (not picking a dime) in each trial \( q = 1 - p = \frac{2}{3} \).
The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]
Where \( \binom{n}{k} \) is the binomial coefficient which represents the number of ways to choose \( k \) successes in \( n \) trials.
1. Calculate the binomial coefficient:
\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3
\]
2. Calculate \( p^k \):
\[
p^k = \left(\frac{1}{3}\right)^2 = \frac{1}{9}
\]
3. Calculate \( q^{n-k} \):
\[
q^{n-k} = \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}
\]
4. Now, plug these values into the binomial probability formula:
\[
P(X = 2) = \binom{3}{2} p^2 q^1 = 3 \times \frac{1}{9} \times \frac{2}{3}
\]
5. Simplify the expression:
\[
P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9}
\]
So, the probability that a dime is picked exactly twice is \( \frac{2}{9} \).
The correct response is:
**Start Fraction 2 over 9 End Fraction**.
- Let \( n = 3 \) (the number of trials).
- Let \( k = 2 \) (the number of successful trials, which is picking a dime).
- The probability of success (picking a dime) in each trial \( p = \frac{1}{3} \) (as there is one dime out of three coins).
- The probability of failure (not picking a dime) in each trial \( q = 1 - p = \frac{2}{3} \).
The binomial probability formula is given by:
\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]
Where \( \binom{n}{k} \) is the binomial coefficient which represents the number of ways to choose \( k \) successes in \( n \) trials.
1. Calculate the binomial coefficient:
\[
\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3
\]
2. Calculate \( p^k \):
\[
p^k = \left(\frac{1}{3}\right)^2 = \frac{1}{9}
\]
3. Calculate \( q^{n-k} \):
\[
q^{n-k} = \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}
\]
4. Now, plug these values into the binomial probability formula:
\[
P(X = 2) = \binom{3}{2} p^2 q^1 = 3 \times \frac{1}{9} \times \frac{2}{3}
\]
5. Simplify the expression:
\[
P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9}
\]
So, the probability that a dime is picked exactly twice is \( \frac{2}{9} \).
The correct response is:
**Start Fraction 2 over 9 End Fraction**.