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]



Answer :

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].