A bag of M\&M's has 2 red, 5 green, 6 blue, and 3 yellow MeM's. What is the probability of randomly picking: (Round to 4 decimat places) a) a yellow? b) a blue or green? c) an orange? Question Help: Message instructor Submit All Parts

**Step 1: Calculate the total number of M&M's** The bag contains: - Red: \(2\) - Green: \(5\) - Blue: \(6\) - Yellow: \(3\) Thus, the total number of M&M's is: \[ \text{Total} = 2 + 5 + 6 + 3 = 16 \] --- **Step 2: Calculate each probability** **a) Probability of picking a yellow M&M:** There are \(3\) yellow M&M's. So, the probability is: \[ P(\text{yellow}) = \frac{3}{16} \approx 0.1875 \] --- **b) Probability of picking a blue or green M&M:** There are \(6\) blue and \(5\) green M&M's. Thus, the total is: \[ 6 + 5 = 11 \] So, the probability is: \[ P(\text{blue or green}) = \frac{11}{16} \approx 0.6875 \] --- **c) Probability of picking an orange M&M:** Since there are no orange M&M's in the bag: \[ P(\text{orange}) = \frac{0}{16} = 0.0000 \] --- **Final Answers:** a) \(0.1875\) b) \(0.6875\) c) \(0.0000\)