Question

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)
a) a yellow?
b) a blue or green?
c) an orange?

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) $0$