At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

A bag held 3 white marbles, 1 orange marble, and 2 green marbles. Anna drew a white marble and did not replace it. What is the probability of drawing a green marble next?

A) [tex]\frac{1}{6}[/tex]
B) [tex]\frac{1}{5}[/tex]
C) [tex]\frac{1}{3}[/tex]
D) [tex]\frac{2}{5}[/tex]


Sagot :

Let's break down the solution to this problem step by step:

1. Initial Count of Marbles:
- White marbles: 3
- Orange marble: 1
- Green marbles: 2

Total initial marbles: \(3 + 1 + 2 = 6\)

2. Marble Drawn:
Anna draws a white marble. Since she does not replace it, we need to update our count:
- Drawn white marbles: 1

3. Remaining Marbles:
After Anna draws one white marble, the counts are as follows:
- White marbles: \(3 - 1 = 2\)
- Orange marble: 1 (no change)
- Green marbles: 2 (no change)

Total remaining marbles: \(2 + 1 + 2 = 5\)

4. Probability Calculation:
We are asked for the probability of drawing a green marble next. The count of remaining green marbles is 2, and the total remaining marbles is 5.

Therefore, the probability \(P\) of drawing a green marble is given by:
[tex]\[ P(\text{green marble}) = \frac{\text{number of green marbles}}{\text{total remaining marbles}} = \frac{2}{5} \][/tex]

Thus, the probability of drawing a green marble next is \(\frac{2}{5}\).

So the correct answer is [tex]\( \boxed{\frac{2}{5}} \)[/tex].