a bag of m&ms has 4 red, 6 green, 3 blue, and 8 yellow m&ms. what is the probability of randomly picking…

a bag of m&ms has 4 red, 6 green, 3 blue, and 8 yellow m&ms. what is the probability of randomly picking (round to 4 decimal places)\na) a yellow?\nb) a blue or green?\nc) an orange?
Answer
Explanation:
Step1: Calculate total number of M&M's
$4 + 6+3 + 8=21$
Step2: Calculate probability of picking a yellow
The number of yellow M&M's is 8. Probability formula is $P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. So $P(\text{yellow})=\frac{8}{21}\approx0.3810$
Step3: Calculate probability of picking a blue or green
The number of blue M&M's is 3 and green is 6. So number of favorable outcomes for blue or green is $3 + 6 = 9$. Then $P(\text{blue or green})=\frac{9}{21}\approx0.4286$
Step4: Calculate probability of picking an orange
There are 0 orange M&M's. So $P(\text{orange})=\frac{0}{21}=0$
Answer:
a) $0.3810$ b) $0.4286$ c) $0$