Answer :
Let's break down the problem step by step:
1. Total Number of Marbles:
The bag contains a total of 15 marbles.
2. Probability of Selecting a Green Marble:
The probability of selecting a green marble is given as \(\frac{1}{5}\).
3. Probability of Selecting a Green Marble and then a Blue Marble:
The probability of selecting a green marble, replacing it, and then selecting a blue marble is given as \(\frac{2}{25}\).
4. Finding the Probability of Selecting a Blue Marble:
Let \(P(G)\) be the probability of selecting a green marble, and \(P(B)\) be the probability of selecting a blue marble. We are given:
[tex]\[ P(G) = \frac{1}{5} \][/tex]
[tex]\[ P(G \text{ and } B) = \frac{2}{25} \][/tex]
Since the events are independent (we replace the green marble before selecting the blue one), we use the multiplication rule for independent events:
[tex]\[ P(G \text{ and } B) = P(G) \times P(B) \][/tex]
Rearranging to solve for \(P(B)\):
[tex]\[ P(B) = \frac{P(G \text{ and } B)}{P(G)} = \frac{\frac{2}{25}}{\frac{1}{5}} = \frac{2}{25} \times \frac{5}{1} = \frac{2 \times 5}{25} = \frac{10}{25} = \frac{2}{5} \][/tex]
5. Number of Blue Marbles:
The probability of selecting a blue marble \(P(B)\) can also be expressed in terms of the number of blue marbles and the total number of marbles:
[tex]\[ P(B) = \frac{\text{Number of Blue Marbles}}{\text{Total Number of Marbles}} \][/tex]
Substituting the known values:
[tex]\[ \frac{2}{5} = \frac{\text{Number of Blue Marbles}}{15} \][/tex]
Solving for the number of blue marbles:
[tex]\[ \text{Number of Blue Marbles} = \frac{2}{5} \times 15 = 6 \][/tex]
Therefore, there are 6 blue marbles in the bag. The correct answer is:
6
1. Total Number of Marbles:
The bag contains a total of 15 marbles.
2. Probability of Selecting a Green Marble:
The probability of selecting a green marble is given as \(\frac{1}{5}\).
3. Probability of Selecting a Green Marble and then a Blue Marble:
The probability of selecting a green marble, replacing it, and then selecting a blue marble is given as \(\frac{2}{25}\).
4. Finding the Probability of Selecting a Blue Marble:
Let \(P(G)\) be the probability of selecting a green marble, and \(P(B)\) be the probability of selecting a blue marble. We are given:
[tex]\[ P(G) = \frac{1}{5} \][/tex]
[tex]\[ P(G \text{ and } B) = \frac{2}{25} \][/tex]
Since the events are independent (we replace the green marble before selecting the blue one), we use the multiplication rule for independent events:
[tex]\[ P(G \text{ and } B) = P(G) \times P(B) \][/tex]
Rearranging to solve for \(P(B)\):
[tex]\[ P(B) = \frac{P(G \text{ and } B)}{P(G)} = \frac{\frac{2}{25}}{\frac{1}{5}} = \frac{2}{25} \times \frac{5}{1} = \frac{2 \times 5}{25} = \frac{10}{25} = \frac{2}{5} \][/tex]
5. Number of Blue Marbles:
The probability of selecting a blue marble \(P(B)\) can also be expressed in terms of the number of blue marbles and the total number of marbles:
[tex]\[ P(B) = \frac{\text{Number of Blue Marbles}}{\text{Total Number of Marbles}} \][/tex]
Substituting the known values:
[tex]\[ \frac{2}{5} = \frac{\text{Number of Blue Marbles}}{15} \][/tex]
Solving for the number of blue marbles:
[tex]\[ \text{Number of Blue Marbles} = \frac{2}{5} \times 15 = 6 \][/tex]
Therefore, there are 6 blue marbles in the bag. The correct answer is:
6
