A bag contains equal numbers of red, green, and yellow marbles. If Geeta pulls three marbles out of the bag, replacing each marble after she picks it. What is the probability that at least one will be red?

Options

A. 7/27
B. 19/27
C. 17/27
D. 13/27

Answer

To solve the problem, we need to find the probability that Geeta pulls at least one red marble when she draws three marbles from a bag containing equal numbers of red, green, and yellow marbles, replacing each marble after each pick.

### Step-by-Step Solution:

1. **Understanding the Total Probability**:
Since the bag contains equal numbers of red, green, and yellow marbles, the probability of drawing a red marble (P(R)) is \( \frac{1}{3} \). Consequently, the probability of not drawing a red marble (P(Not R)) is \( \frac{2}{3} \).

2. **Finding the Probability of No Red Marbles**:
We need to calculate the probability that none of the three marbles drawn are red. Since the draws are independent (because the marbles are replaced after each draw), we can multiply the probabilities for each draw:
\[ P(\text{No Red in 3 draws}) = P(\text{Not R}) \times P(\text{Not R}) \times P(\text{Not R}) = \left(\frac{2}{3}\right)^3 \]
\[ = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \frac{8}{27} \]

3. **Calculating the Probability of At Least One Red Marble**:
The probability of getting at least one red marble is the complement of the probability of getting no red marbles. Therefore:
\[ P(\text{At least one Red}) = 1 - P(\text{No Red in 3 draws}) = 1 - \frac{8}{27} \]
\[ = \frac{27}{27} - \frac{8}{27} = \frac{19}{27} \]

### Final Answer:
The probability that at least one of the marbles drawn is red is \( \frac{19}{27} \).