Question

A bag of M&M's has 2 red, 3 green, 8 blue, and 6 yellow M&M's. What is the probability of randomly picking: (Round to 4 decimal places)

Answer 1

Given:

A bag of M&M's has 2 red, 3 green, 8 blue, and 6 yellow M&M's.

Total M&M's = 2+3+8+6=19

a) The probability of randomly picking a yellow M&M's is,

$\begin{gathered} P=\frac{Number\text{ of possible outcomes}}{\text{Total number of outcomes}} \\ P=\frac{6}{19}=0.3158 \end{gathered}$

b) The probability of randomly picking a blue or green M&M's is,

$P=\frac{8}{19}+\frac{3}{19}=\frac{11}{19}=0.5789$

c) The probability of randomly picking a orange M&M's is,

As there are no orange M&M's in the bag, the probability is,

$P=\frac{0}{19}=0$