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. You randomly select peanut M&M’s from an extra-large bag looking for a green candy.

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Melanie 3 weeks 2021-11-10T16:08:37+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-11-10T16:09:45+00:00

    Answer:

    The questions asked are

    If you randomly select 4 peanuts

    1. Compute the probability that exactly three of the four M&M’s are brown

    2. Compute the probability that two or three of the four M&M’s are brown.

    3. Compute the probability that at most three of the four M&M’s are brown.

    4. Compute the probability that at least three of the four M&M’s are brown.

    Step-by-step explanation:

    Given the following information

    Brown=12%. P(B)=0.12

    Yellow=15%. P(Y)=0.15

    Red=12%. P(R), =0.12

    Blue=23%. P(B) =0.23

    Orange, =23%. P(O) =0.23

    Green=15%. P(G)=0.15

    Question 1.

    They are independent events

    If there are exactly three brown and the last is not brown

    P(B n B n B n B’)

    P(B)×P(B)×P(B)×P(B’)

    0.12×0.12×0.12×(1-P(B))

    0.001728×(1-0.12)

    0.001728×0.88

    0.00152.

    0.152%

    2. If two or three are brown

    I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B’)×P(B’))+ (P(B)×P(B)×P(B’))

    (0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

    0.0112+0.00152

    0.0127

    1.27%

    3. At most three brown out of four then we are going to have

    BBBB’, BBB’B’, BB’B’B’, B’B’B’B’

    These are the cases of at most three brown.

    P(B)×P(B)×P(B)×P(B’) + P(B)×P(B)×P(B’)×P(B’) + P(B)×P(B’)×P(B’)×PB’)+ P(B’)×P(B’)×P(B’)×P(B’)=

    0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

    0.694

    69.4%

    4. At least 3 brown out of four selection

    I.e BBBB’, BBBB

    These are the two options

    P(B)×P(B)×P(B)×P(B’) + P(B)×P(B)×P(B)×P(B)=

    0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

    0.001728

    0.173%

    0
    2021-11-10T16:10:03+00:00

    Answer:

    The probability that the first green candy is the seventh M&M selected = 0.0566

    Step-by-step explanation:

    Probability of selecting a green candy = P(G) = 15% = 0.15

    Probability of not selecting a green candy, P(G’) = 1 – 0.15 = 0.85

    To compute the probability that the first green candy is the seventh M&M selected

    For this to happen, the not green candies are selected 6 times, before a green one is drawn on the 7th draw

    That is

    [P(G’)]⁶ (P(G) = 0.85⁶ × 0.15 = 0.05657 = 0.0566

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