To determine the probability that the sum of two coins picked is at least 30 cents, we need to consider the potential outcomes when Kevin randomly picks a coin twice, with replacement.
1. Identify the value of each coin:
- Dime: 10 cents
- Nickel: 5 cents
- Quarter: 25 cents
2. Calculate the probability of picking each type of coin:
Since Kevin has an equal number of dimes, nickels, and quarters, the probability [tex]\( P \)[/tex] of picking any one coin (dime, nickel, or quarter) is the same and is given by:
[tex]\[
P_{\text{dime}} = P_{\text{nickel}} = P_{\text{quarter}} = \frac{1}{3}
\][/tex]
3. Enumerate the combinations that result in a sum of at least 30 cents:
- Two quarters: [tex]\( 25 + 25 = 50 \)[/tex] cents
- One quarter and one dime: [tex]\( 25 + 10 = 35 \)[/tex] cents
- One dime and one quarter: [tex]\( 10 + 25 = 35 \)[/tex] cents
4. Calculate the probability of each favorable combination:
- Probability of picking two quarters:
[tex]\[
P(25 + 25) = P_{\text{quarter}} \times P_{\text{quarter}} = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9}
\][/tex]
- Probability of picking a quarter first and then a dime:
[tex]\[
P(25 + 10) = P_{\text{quarter}} \times P_{\text{dime}} = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9}
\][/tex]
- Probability of picking a dime first and then a quarter:
[tex]\[
P(10 + 25) = P_{\text{dime}} \times P_{\text{quarter}} = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9}
\][/tex]
5. Determine the total probability for the desired event:
Sum the individual probabilities of the favorable outcomes:
[tex]\[
\text{Total probability} = P(25 + 25) + P(25 + 10) + P(10 + 25) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3}
\][/tex]
The probability that the sum of the two coins picked is at least 30 cents is:
[tex]\[
\boxed{\frac{1}{3}}
\][/tex]
Thus, the correct answer is [tex]\( D \)[/tex]:
[tex]\[
\boxed{\frac{1}{3}}
\][/tex]
