A bag of M&M's has 6 red, 5 green, 4 blue, and 8 yellow M&M's. What is the probability of randomly picking: (give answer

Question

2 years ago

12

as a reduced fraction) 1) a yellow? w 2) a blue or green? 3) an orange?

2 answers:

Dahasolnce [82] · Answer 1 · 2022-05-14T18:01:29+03:00

6 0

Answer:

P( yellow) = 8/23

P( blue or green) = 9/23

P(orange) = 0

Step-by-step explanation:

6 red, 5 green, 4 blue, and 8 yellow M&M's = 23 total

P( yellow) = yellow / total = 8/23

P( blue or green) = (blue+green) / total = (5+4)/23 = 9/23

P(orange) = orange/ total = 0/23

kenny6666 [7] · Answer 2 · 2022-05-14T20:08:25+03:00

3 0

Answer:

there are 23 m&m's.

Step-by-step explanation:

Probability of getting red is 6/23

Probability of getting green is 5/23

Probability of getting blue is 4/23

Probability of getting yellow is 8/23

Orange = red + yellow = 6+8/23

Probability of getting Orange = 14/23