Question

2 each point on the graph below represents a different kind of soda. what can we say about point b? i have a bag of 10 m&ms. in the bag there are 2 yellows, 2 browns, 3 reds, and 3 greens. 3 what is the probability of randomly choosing a red m&m? 4 what is the probability of choosing a red or a green m&m? 5 what is the probability of randomly choosing a yellow m&m, eating it, and then choosing another yellow m&m?

Explanation:

Question 3:

Step1: Identify total and red count

Total M&Ms: \( 2 + 2 + 3 + 3 = 10 \)
Red M&Ms: \( 3 \)

Step2: Calculate probability

Probability \( = \frac{\text{Red count}}{\text{Total count}} = \frac{3}{10} \)

Step1: Identify red and green counts

Red M&Ms: \( 3 \), Green M&Ms: \( 3 \)
Total M&Ms: \( 10 \) (from Q3)

Step2: Calculate OR probability (non - overlapping)

Since red and green are distinct, \( P(\text{Red OR Green}) = \frac{\text{Red} + \text{Green}}{\text{Total}} = \frac{3 + 3}{10} = \frac{6}{10}=\frac{3}{5} \)

Step1: First choice (yellow)

Yellow M&Ms: \( 2 \), Total: \( 10 \)
Probability of yellow: \( \frac{2}{10}=\frac{1}{5} \)

Step2: Second choice (yellow, after eating one)

After eating one yellow, yellow left: \( 1 \), Total left: \( 9 \)
Probability of second yellow: \( \frac{1}{9} \)

Step3: Multiply probabilities (AND event)

\( P(\text{Yellow then Yellow}) = \frac{2}{10} \times \frac{1}{9}=\frac{2}{90}=\frac{1}{45} \)

Answer:

\( \frac{3}{10} \)

Question 4: