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

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)=(11)/(100)+(21)/(100) \\ =(32)/(100) \\ =(8)/(25) \end{gathered}`

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

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

(3) The probability that they would both be orange

`\begin{gathered} Pr(YY)=(21)/(100)*(21)/(100) \\ =(441)/(10000) \end{gathered}`