Question

each package of M&M’s should contain 24% blue, 13% brown, 15% green, 28% orange, 12% red, and 18% yellow M&M’s on average. If you chose a random M&M, what is the probability that it would be red or blue?If you chose a random M&M, what is the probability that it wouldn’t be yellow?If you chose 2 M&M’s, what is the probability that they would both be orange?

Answer 1

Solution

Blue = 21%

Brown = 14 %

green = 18%

orange =21 %

red = 11%

yellow = 15%

Total = 100 %

(1)

The probability that it would be red or blue?

$\begin{gathered} Pr(RorB)=\frac{11}{100}+\frac{21}{100} \\ =\frac{32}{100} \\ =\frac{8}{25} \end{gathered}$

(2) The probability that it wouldn’t be yellow?

$\begin{gathered} 1-Pr(yellow) \\ =1-\frac{15}{100} \\ =\frac{85}{100} \\ =\frac{17}{25} \end{gathered}$

(3) The probability that they would both be orange

$\begin{gathered} Pr(YY)=\frac{21}{100}\times\frac{21}{100} \\ =\frac{441}{10000} \end{gathered}$