Question

A jawbreaker candy machine has 25 red jawbreakers, 30 green jawbreakers, 15 yellow

Answer 1

Answer: Choice C, 2/3

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Work Shown:

A = # of candies that are not green

A = (# of red)+(# of yellow)+(#  of blue)

A = 25+15+20

A = 60

B = # of candies total

B = (# of red)+(# of green)+(# of yellow)+(# of blue)

B = 25+30+15+20

B = 90

P(not green) = probability of selecting candy that is not green

P(not green) = (# of candies not green)/(# of candies total)

P(not green) = A/B

P(not green) = 60/90

P(not green) = (2*30)/(3*30)

P(not green) = 2/3