A bag contains 8 red M&Ms, 12 blue M&Ms, 3 brown M&Ms, and 7 green M&Ms. An M&M is picked at random from the bag.

Question

A bag contains 8 red M&Ms, 12 blue M&Ms, 3 brown M&Ms, and 7 green M&Ms. An M&M is picked at random from the bag.

What is the probability of getting the first color M&M, eating it, then grabbing the second color?

Brown, then blue:

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Dulcie 2 months 2021-07-29T10:46:07+00:00 1 Answers 1 views 0

Answers ( )

    0
    2021-07-29T10:47:26+00:00

    Answer:

    4%

    Step-by-step explanation:

    8 red + 12 blue + 3 brown + 7 green = 30 M&Ms

    3 brown / 30 M&Ms = 0.1

    you ate one so now you have 29 M&Ms

    12 / 29 = 0.4137931034

    0.1 x 0.4137931034 x 100 = 4.1379310344%

    rounds to 4%

    hope this helped

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