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

Respuesta :

wegnerkolmp2741o

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

dipo142008

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

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