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 b

Question

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 a randomly selected peanut M&M is not yellow.
(b) Compute the probability that a randomly selected peanut M&M is orange or yellow.
(c) Compute the probability that three randomly selected peanut M&M’s are all red.
(d) If you randomly select two peanut M&M’s, compute that probability that neither of them are red.
(e) If you randomly select two peanut M&M’s, compute that probability that at least one of them is red.

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1 month 2021-09-14T00:48:58+00:00 1 Answer 0

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    2021-09-14T00:50:02+00:00

    Answer:

    (a)0.85

    (b)0.38

    (c)0.002

    (d)0.986

    (e)0.12

    Step-by-step explanation:

    Brown = 12%,

    Yellow = 15%

    Red = 12%

    Blue = 23%

    Orange= 23%

    Green = 15%

    (a) Probability that a randomly selected peanut M&M is not yellow.

    P(Yellow) = 15/100 =0.15

    Therefore: P(Not Yellow)

    =1-0.15 =0.85

    (b) Probability that a randomly selected peanut M&M is orange or yellow.

    P(Orange OR Yellow)

    = P(Orange)+P(Yellow)

    =23/100 + 15/100

    =38/100 =0.38

    (c) Probability that three randomly selected peanut M&M’s are all red.

    P(Red and Red and Red) = 12/100 X 12/100 X 12/100 =0.001728 = 0.002

    (d) Probability that neither of them are red.

    P(neither of them are red)= 1- P(both are red) = 1-(12/100 X 12/100)

    = 1-0.0144 = 0.9856 = 0.986

    (e)Probability that at least one of them is red.

    P(RB or RY or RR or Rb or RO or RG)

    =(0.12 X 0.12)+(0.12 X 0.15)+(0.12 X 0.12)+(0.12 X 0.23)+(0.12 X 0.23)+ (0.12 X 0.15)

    = 0.12

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