Question

Theoretical Probabilities of Repeated Events Quick Check
5 of 55 of 5 Items
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

19
Start Fraction 1 over 9 End Fraction

727
Start Fraction 7 over 27 End Fraction

29
Start Fraction 2 over 9 End Fraction

23

Answer 1

To solve this problem, we can use the binomial probability formula since the draws are independent, and the coin is replaced after each draw.

The probability of picking a dime (\(p\)) is \( \frac{1}{3} \) since there are three coins (nickel, penny, and dime). The probability of not picking a dime (\(q\)) is \(1 - p = \frac{2}{3}\).

We want to find the probability of picking a dime exactly 2 times in 3 trials. The binomial probability formula is given by:

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

where:
- \(n\) = total number of trials (3)
- \(k\) = number of successful outcomes (2)
- \(p\) = probability of success (picking a dime)
- \(q\) = probability of failure (not picking a dime)

Calculating:

1. \(\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3\)
2. \(p^k = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)
3. \(q^{n-k} = \left(\frac{2}{3}\right)^{3-2} = \left(\frac{2}{3}\right)^1 = \frac{2}{3}\)

Now plug these values into the formula:

\[
P(X = 2) = 3 \times \frac{1}{9} \times \frac{2}{3} = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9}
\]

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

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

The correct response is:

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