Let’s find the probability of drawing three specific coins in sequence: a nickel first, a dime second, and a quarter third, from a collection of four nickels, three dimes, and three quarters.
1. Total number of coins:
- Nickels: 4
- Dimes: 3
- Quarters: 3
[tex]\[ \text{Total coins} = 4 + 3 + 3 = 10 \][/tex]
2. Probability of picking the first coin as a nickel:
There are 4 nickels out of 10 coins.
[tex]\[ P(\text{Nickel first}) = \frac{4}{10} \][/tex]
3. Probability of picking the second coin as a dime:
After picking one nickel, we have 9 coins left, including 3 dimes.
[tex]\[ P(\text{Dime second} \mid \text{Nickel first}) = \frac{3}{9} = \frac{1}{3} \][/tex]
4. Probability of picking the third coin as a quarter:
After picking one nickel and one dime, we have 8 coins left, including 3 quarters.
[tex]\[ P(\text{Quarter third} \mid \text{Nickel first and Dime second}) = \frac{3}{8} \][/tex]
5. Combined probability of the sequence:
The combined probability, using the multiplication rule for dependent events, is:
[tex]\[ P(\text{Nickel first, Dime second, Quarter third}) = P(\text{Nickel first}) \times P(\text{Dime second} \mid \text{Nickel first}) \times P(\text{Quarter third} \mid \text{Nickel first and Dime second}) \][/tex]
Substituting the probabilities found:
[tex]\[ P(\text{Nickel first, Dime second, Quarter third}) = \frac{4}{10} \times \frac{1}{3} \times \frac{3}{8} \][/tex]
6. Simplifying the probability:
[tex]\[ P(\text{Nickel first, Dime second, Quarter third}) = \frac{4}{10} \times \frac{1}{3} \times \frac{3}{8} = \frac{4 \times 1 \times 3}{10 \times 3 \times 8} = \frac{12}{240} = \frac{1}{20} \][/tex]
Therefore, the probability of drawing a nickel first, a dime second, and a quarter third is:
[tex]\[ \boxed{\frac{1}{20}} \][/tex]
The correct answer is:
D. [tex]\(\frac{1}{20}\)[/tex]
