\( \begin{array}{r}\text { A bag has } 20 \text { cubes in it. } 6 \text { of the cubes are green. } \\ \text { You take one cube out of the bag at random. }\end{array} \) \( \begin{array}{lllll}\frac{6}{14} & 30 \% & 0.6 & \frac{3}{10} \\ \text { Which four values below show the probability that you take out a cube that is green? }\end{array} \) \( \begin{array}{lllll}6 \% & \frac{3}{5} & \frac{6}{20} & 0.03 \\ 0.3 & \frac{6}{10} & 60 \% & \frac{6}{26}\end{array} \) Q zoom

To determine the probability of drawing a green cube from the bag, let's break down the problem: - **Total cubes in the bag:** 20 - **Green cubes:** 6 **Probability Formula:** \[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{6}{20} \] Simplifying the fraction: \[ \frac{6}{20} = \frac{3}{10} = 0.3 = 30\% \] So, the probability of drawing a green cube can be represented in four different ways: 1. \(\frac{6}{20}\) 2. \(\frac{3}{10}\) 3. \(0.3\) 4. \(30\%\) **From the provided options:** - **First Row:** - \(6/14\) ➔ **Incorrect** - **30\%** ➔ **Correct** - \(0.6\) ➔ **Incorrect** - **\(\frac{3}{10}\)** ➔ **Correct** - **Second Row:** - \(6\%\) ➔ **Incorrect** - \(3/5\) ➔ **Incorrect** - **\(6/20\)** ➔ **Correct** - \(0.03\) ➔ **Incorrect** - **\(0.3\)** ➔ **Correct** - \(6/10\) ➔ **Incorrect** - \(60\%\) ➔ **Incorrect** - \(6/26\) ➔ **Incorrect** **Therefore, the four correct values are:** - \(30\%\) - \(\frac{3}{10}\) - \(6/20\) - \(0.3\) **Answer:** The four correct probabilities are 30 %, 3⁄10, 6⁄20, and 0.3.