SOLUTION: According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. [Roun

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Question 1134928: According to Masterfoods, the company that manufactures M&M’s, 12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. [Round your answers to three decimal places, for example: 0.123]
Compute the probability that two randomly selected peanut M&M’s are both brown.

If you randomly select two peanut M&M’s, compute that probability that neither of them are orange.


If you randomly select two peanut M&M’s, compute that probability that at least one of them is orange.

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
According to Masterfoods, the company that manufactures M&M’s,
12% of peanut M&M’s are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green.
[Round your answers to three decimal places, for example: 0.123]

(a)  Compute the probability that two randomly selected peanut M&M’s are both brown.


         P = 0.12*0.12 = 0.0144 = 0.014 (rounded as requested).    ANSWER

             ( 0.12 = 12% )



(b)  If you randomly select two peanut M&M’s, compute that probability that neither of them are orange.


         P = (1-0.12)*(1-0.12) = 0.88*0.88 = 0.7744 = 0.774 (rounded as requested).    ANSWER


             (My calculations should be clear to you without further explanations !)



(c)  If you randomly select two peanut M&M’s, compute that probability that at least one of them is orange.


         P = 0.23*0.23 + 0.23*(1-0.23)


             This time  YOU  make all necessary calculations and rounding on your own.


             Happy calculations !


Who does not like  M&M’s  ?  ?     (rhetoric question . . . )


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