Answer :
Certainly! Let's go step-by-step to solve this problem:
1. Initial Count of Marbles:
- Red marbles: 2
- Blue marbles: 3
- Yellow marbles: 5
The total number of marbles initially is:
[tex]\[ 2 + 3 + 5 = 10 \][/tex]
2. Initial Number of Blue Marbles:
Initially, there are 3 blue marbles.
3. Change After Drawing a Blue Marble:
Marco drew one blue marble and did not replace it. So,
- The total number of marbles now is:
[tex]\[ 10 - 1 = 9 \][/tex]
- The number of blue marbles now is:
[tex]\[ 3 - 1 = 2 \][/tex]
4. Probability of Drawing a Blue Marble Next:
The probability is calculated as the number of favorable outcomes (drawing a blue marble) divided by the total number of possible outcomes (remaining marbles):
[tex]\[ \text{Probability} = \frac{\text{Number of Blue Marbles}}{\text{Total Number of Marbles}} = \frac{2}{9} \][/tex]
So, the probability of drawing a blue marble next is \(\frac{2}{9}\).
The correct answer is:
[tex]\[ B) \frac{2}{9} \][/tex]
1. Initial Count of Marbles:
- Red marbles: 2
- Blue marbles: 3
- Yellow marbles: 5
The total number of marbles initially is:
[tex]\[ 2 + 3 + 5 = 10 \][/tex]
2. Initial Number of Blue Marbles:
Initially, there are 3 blue marbles.
3. Change After Drawing a Blue Marble:
Marco drew one blue marble and did not replace it. So,
- The total number of marbles now is:
[tex]\[ 10 - 1 = 9 \][/tex]
- The number of blue marbles now is:
[tex]\[ 3 - 1 = 2 \][/tex]
4. Probability of Drawing a Blue Marble Next:
The probability is calculated as the number of favorable outcomes (drawing a blue marble) divided by the total number of possible outcomes (remaining marbles):
[tex]\[ \text{Probability} = \frac{\text{Number of Blue Marbles}}{\text{Total Number of Marbles}} = \frac{2}{9} \][/tex]
So, the probability of drawing a blue marble next is \(\frac{2}{9}\).
The correct answer is:
[tex]\[ B) \frac{2}{9} \][/tex]
