Question

juan has 3 dimes, 1 penny, 3 nickels, and 2 quarters in his pocket. he randomly selects three coins with replacement. which expression represents the probability that the first coin is a penny, and the second and third coins are dimes?\\(\big(\frac{1}{9}\big)^3\\)\\(\frac{1}{9}\cdot\frac{3}{9}\cdot\frac{3}{9}\\)\\(\big(\frac{1}{4}\big)^3\\)\\(\frac{1}{9}\cdot\frac{3}{9}\cdot\frac{2}{9}\\)\\(\frac{1}{9}+\frac{3}{9}+\frac{3}{9}\\)

Explanation:

Step1: Calculate total number of coins

First, find the total number of coins. Juan has 3 dimes, 1 penny, 3 nickels, and 2 quarters. So total coins = 3 + 1 + 3 + 2 = 9.

Step2: Probability of first coin (penny)

The probability of selecting a penny first: there is 1 penny out of 9 coins, so probability is $\frac{1}{9}$.

Step3: Probability of second coin (dime)

Since we are selecting with replacement, the total number of coins remains 9. The number of dimes is 3, so probability of selecting a dime is $\frac{3}{9}$.

Step4: Probability of third coin (dime)

Again, with replacement, total coins are 9 and dimes are 3, so probability of selecting a dime is $\frac{3}{9}$.

Step5: Combine probabilities

Since the events are independent (because of replacement), we multiply the probabilities of each event. So the probability is $\frac{1}{9} \cdot \frac{3}{9} \cdot \frac{3}{9}$.

Answer:

$\frac{1}{9} \cdot \frac{3}{9} \cdot \frac{3}{9}$ (the second option: $\frac{1}{9} \cdot \frac{3}{9} \cdot \frac{3}{9}$)